1538240417592_optics.pdf

PHYSICS OPTICS

DISPERSION Dispersion Light : When a beam of composite light (consisting of several wavelength) passes through a prism, it splits into its constituent colours. This phenomenon is called dispersion and the band of colour obtained on a screen is called the spectrum. If white light is used, seven colours are obtained.

Production of pure spectrum : Newton’s spectrum of sunlight is an impure spectrum because the different coloured images overlap. A pure spectrum is one in which the different coloured images contain light of one colour only, i.e., they are monochromatic images. In order to obtain a pure spectrum. i) the white light must be admitted through a very narrow opening, so as to assist in the reduction of the overlapping of the images. ii) the beams of coloured rays emerging from the prism must be parallel, so that each beam can be brought to a separate focus.

The spectrometer can be used to provide a pure spectrum. The collimator slit is made very narrow and the collimator C and the telescope T are both adjusted for parallel light. A bright source of white light S is placed near the slit and the prism P is usually set in the minimum deviation position for yellow light, although this is not essential. The ray refracted through P are now separated into a number of different coloured parallel beams of light each travelling in slightly different directions, and the telescope bring each coloured beam to a separate focus. A pure spectrum can now be seen through T, consisting of a series of monochromatic images of the slit. Types of spectra : Spectra are of two types - a) Emission and (b) Absorption spectra. Emission spectra : Spectra obtained from luminous bodies are called emission spectra. Such spectra are of three types. i) Line spectrum : A line spectrum consists of narrow bright lines separated by dark intervals. Line spectra are emitted by substances in the atomic state and are characteristic of the substances emitting them. ii) Band spectrum : A band spectrum consists of a number of bright separated by dark intervals. Band spectra are emitted by substances in the molecular state. A band consists of a large number of close lines.

PHYSICS iii) Continuous spectrum: A continuous spectrum consists of an unbroken sequence of wavelengths (or colours) over a wide range. Continuous spectra are emitted by solids, liquids and highly compressed gases heated to high temperature. A continuous spectrum is not characteristic of the nature of the source but depends only on the temperature. Absorption spectra : When white light passes through a semi-transparent solid, liquid or gas, its spectrum contains certain dark lines or bands showing that certain wavelengths have been absorbed. Such a spectrum is called the absorption spectrum of the substances through which light is passed. Substances in atomic state produce line absorption spectra and substances in molecular state produce band absorption spectra. A substance in vapour state absorbs light of the same wave lengths as it would emit at the same temperature. For example, if white light is made to pass through sodium vapour, the continuous spectrum will have two dark lines corresponding to the two yellow lines of sodium. Fraunhofer lines : The solar spectrum shows several dark lines crossing the other continuous spectrum. These are called Fraunhofer lines. The central part of the sun, called the photosphere, which is at very high temperature ( ≈ 107°C ) emits white light. When this light passes through the chromospheres, which is at a much lower temperature ( 6000°C ), the gases absorb light of certain wavelengths, resulting in a reduced intensity of these wavelengths. The study of Fraunhofer lines has given information about the presence of various elements in the atmosphere of the sun. Atmospheric refraction : The apparent shift in the position of the sun at sunrise and sunset: This effect is due to atmospheric refraction of light and can be explained as below:

The line XY represents the horizon for an observer O on the surface of the earth. Near the surface of the earth air is a little denser than outer atmosphere and its refractive index is 1.00029 w.r.t outer space. Therefore, a ray of light travelling from the sun towards earth, travels from outer space to air, which is optically a little denser. Due to this, it undergoes refraction and therefore deviates towards the normal. To the observer O on the surface of earth, the sun appears at S′. The apparent shift in the direction of the sun is only 0.5° and the time difference between actual sunset and the apparent sunset is about 2 minutes. The apparent flattening of the disc of the sun at sunset and sunrise is also due to the same reason. Twinkling of stars : The refractive index of atmosphere changes with height. Even at the same level, the refractive index of air varies periodically. The rays of light from a star are sometimes concentrated at a point when it appears bright, next moment the concentration of rays decreases and the star appears faint. The planets being nearer, the amount of light received from them is greater and so the variation of brightness is not appreciable.

PHYSICS Scattering of light in atmosphere : Blue colour of sky : When light from the sun travels through earth’s atmosphere, it gets scattered by the large number of molecules of various gases. It is found that the amount of scattering by molecules, called ray light scattering is inversely proportional to the fourth power of the wavelength. Thus light of shorter wavelengths is scattered much more than the light of longer wavelengths since blue colour has relatively shorter wavelength, it predominates and sky appears bluish. Colour of clouds: Large particles like water droplets and dust do not have this selective scattering power. They scatter wavelengths almost equally. Hence clouds appear to be white. Colour of the sun at sunrise and sunset: At sunrise or sunset, the sun is near horizon. Light from the sun therefore has to pass through a larger thickness of the atmosphere than to noon when the sun is overhead. Due to this, more of the blue light is scattered away the sun appears reddish at sunrise and sunset. Optical Instrument Simple microscope: A single convex lens behaves as a simple microscope. If the object (AB) is placed within focus of a convex lens, then the convex lens forms its magnified and erect images on the same side as that of object.

Magnifying power of this microscope is Angle subtended by image at eye (β) M=  Angle subtended by the object ( AB ) when it is supposed to be placed at    least distance for distinct vision or in the position of final image A B α ( )( ) 1 2   AB AB tan β A1 B . A1 L ∴ M= = = 1 1= 1 1 tan α A1 L . A1 B2 A1 B2 AB

In similar triangles A1 B1 L and ABL,

A1 B1 v = AB u

∴ M =

v u

1 1 1 = − f v u Both u and v are negative, we have 1 1 1 v v v D =− + or, = +1 or, M = 1 + = 1 + where D is the least distance for distinct f v u u f f f vision. Since D is fixed, ‘f ’ should be small for large magnifying power. Compound microscope : It consists of a long cylindrical metallic tube carrying at one end an achromatic convex lens O of small focal length, and small aperture. This is called objective lens. At the other end of this tube is fitted an achromatic convex lens E whose focal length and aperture are more than that of objective. Cross-wires are mounted on the focus of the eyepiece. But,

PHYSICS Formation of Image : The object AB lies just outside the focus of objective O which forms its real, bigger and inverted image at A1 B1 . This image lies between the pole and focus of the eyepiece and forms its virtual

inverted image A2 B2 . However if the image A1 B1 lies at the focus of eyepiece, then the final image will be formed at infinity (relaxed eye). Magnifying power : Angle subtended by image at eye ( β ) M= Angle subtended by the object at eye when the object is

supposed to be placed at the position of the finalimage ( α ) ∴

M=

β tan β A1 B1 ( ve ) = = α tan α ( −ue ) AB

But in similar triangles ABO and A1 B1O, Therefore M = −

Therefore,

v0 ve . . u0 ue

1 1 1 = + f e ve ue

But

or,

A1 B1 v0 = AB u0

1 1 1 = − , f e ve ue

1 1 1 + = f e ve ue

ve and ue are both negative.

or,

ve v = 1+ e ue fe

ve  D  v  D = 1 +  or, M = − 0 + 1 +  and the length of microscope is L = ve + ue ue  fe  u0  fe  However, if the final images is formed at infinity, then ve = D but ue = fe But ve = D

∴

Therefore M = −

v0 D and the length of microscope is L = ve + f e . u0 f e

Note: To increase the magnifying power of the microscope : a) Focal length of objective should be small. b) Object should be placed very near to first focus of objective. c) Focal length of eyepiece should be small. Magnifying power of telescopes (angular magnification) : Angular magnification of telescopes is defined as Anglesubtended by image at eye ( β ) M= Anglesubtended by the object at eye when the object is in its actual position ( α )

PHYSICS Astronomical telescope: Construction: It consists of a long cylindrical metallic tube carrying an achromatic convex lens of large focal length and large aperture. This lens faces the object therefore it is called the ‘objective lens.’ At the other end of the tube a smaller tube is fitted which can be moved in and out of the bigger tube with the help of rack and pinion arrangement. At the other end of the smaller tube is fitted an achromatic convex lens of small focal length and small aperture. This lens faces the eye, therefore it is known as eyepiece. Cross wires are mounted in the smaller tube and at the focus of the eyepiece.

Formation of image : Figure shows the objective lens L1 and eyepiece L2 of the telescope. The lens

L1 forms a small, inverted, real image of the distant object AB on its own second focus Fo . This image lies within first focus Fe of the eyepiece. This image A1 B1 acts as an object for the eyepiece and the eyepiece forms a virtual, erect ( w.r.t. A1 B1 ) and magnified final image A2 B2 . Magnifying power : The angular magnification Anglesubtended byimage at eye ( β ) M= Anglesubtended by object at eye when the object is in its actual position ( α )

Since, β and α are small, tan β = β; tan α = α M=

f tan β A1 B1 × A1 L1 A1 L1 = = = 0 tan α A1 L2 × A1 B1 A1 L2 −ue

This is the general formula for magnifying power. Now two cases arise : i) When the final image is formed at least distance of distinct vision: If the distance of the final 1 1 1 image A2 B2 from the eyepiece be D, than while applying the lens formula, − = v u f For the eyepiece we shall have v = − D, u = −ue and f = + fe ∴

1 1 1 =− − fe D −ue

or,

Substituting this value of

f  1 1 1 1 = + = 1 + e  ue f e D f e  D f  f  1 in equation (i), we get, M = − 0 1 + e  ue fe  D

Length of the telescope in this position is ‘L’ = f 0 + ue ii) When the final image is formed at infinity: To see with relaxed eye the final image should be formed at infinity. For this, the distance between the eyepiece and the objective is adjusted so that the image A1 B1 formed by the objective O is at the focus Fe of the eyepiece ( ue = f e ) . This adjustment of the telescope is called ‘normal adjustment’.

PHYSICS

Substituting ue = fe in equation (i), we get M = − f 0 f e

In this position the length of the telescope is maximum and is given by L = f 0 + fe . HUMAN EYES : Power of accommodation : The purpose of the lens of a human eye is to form a real image on the retina. In order to focus objects located at different distances, the focal length of the lens of the eye changes by muscular effects. The ability of the eye lens to adjust its own focal length is known as accommodation. For close vision muscles have to compress the edge of the eye lens so that the lens may become a thick lens of smaller focal length.

For the normal eye, far point is at infinity and near point is at D = 25 cm In doing so the muscles are in tension. This explains why a normal eye cannot see comfortably objects closer than certain limit known as least distance of distinct vision. The distance is usually 25 cm for a normal eye. Thus the near-point for a normal eye is at about 25 cm. When the muscles are relaxed, the eye lens has small curvature and large focal length. A normal eye can see distant objects without any strain. In fact the far-point for a normal eye is at infinite distance. A normal eye can see objects over wide range of distances from 25 cm to infinity, but a defective eye is not able to see objects for such a wide range of distance. Adaption : As known to everybody, the function of the pupil is to control the quantity of light entering the eye. It involuntarily contracts or dilates according as the external light is strong or weak. This process is known as adaptation. The contraction or dilation of the pupil is, however, not instantaneous; it requires a few seconds. At night when the electric light is suddenly switched off, a person in the room feels almost blind, but gradually he recovers the sensation of vision. This is because, when the electric light was on, the pupil was small. When the light is put off, generally a small quantity of light still enters the room through doors, windows etc. But the pupil being small, very small quantity of light still enters the eye to produce the sensation of vision and the person feels blind. But gradually the pupil dilates and after sometimes it becomes fairly large so that more light enters the eye and the person sees almost everything in the room. Similarly a person going out of a dark room into bright sunlight feels dazzled for a moment, gradually his pupils contract and he sees things clearly.

PHYSICS Defect of vision : A normal eye by virtue of accommodation can focus parallel rays on the retina and can see distinctly an object at any position from a very great distance to the nearest point of distinct vision. As eye which fails to do this is called defective The two common defects are (i) short sight or myopia and (ii) long sight or hypermetropia. Besides there are other two defects. e.g. (iii) presbyopia and (iv) astigmatism. The first three defects involve displacement of the near or the far point or both. Myopia : When an eye cannot see objects clearly beyond certain distance, the eye is said to be short sighted or myopic. This happens when the focal length of the eye-lens becomes too short or the eyeball becomes much elongated. Under this condition, parallel rays of light coming from distant object meet at a point in front of the retina, and hence the distant objects cannot be seen clearly. The far point of myopic eye is nearer than infinity, usually quite close to the eye. The nearer point may be within a few centimetre of the eye.

Remedy : To see a distant object its virtual image must be brought to the far point F of the eye (fig 2). The fig 1 shows a myopic eye. The eye-lens due to elongation is more converging than necessary. This defect can be removed by using a concave lens (diverging lens), with its principal focus at F must be used close to the eye (fig 2). A short sighted person whose far point is nearer to the eye than 25cm must use glasses even for reading such glasses must also be divergent. Hypermetropia : When eye cannot see near objects distinctly, it is said to be long-sighted or hypermetropic. When the focal length of the eye lens becomes longer or the eye-ball becomes shortened, hypermetropia develops. Under this condition parallel rays of light after refraction through the eye lens meet at a point behind the retina. Distant objects may be seen by exerting accommodation, while near objects cannot be seen clearly. Unless a beam is convergent a hypermetropic eye cannot without accommodation bring it to a focus on retina. The far point is thus virtual. It may also happen that the eye is altogether incapable of bringing any divergent bundle to a focus on the retina. The near point is then also virtual and lies behind the retina.

Remedy: The fig 1 shows the condition of a long-sighted eye. This defect is due to the reflecting system of the eye being too weak for the cornea-to-retina distance. It can be corrected by strengthening the refraction with the help of a convex lens (fig 2) of suitable focal length so that parallel rays of light after passing through the lens may form image on the retina. If the near point of hypermetropic eye is d cm away from eye, the lens must be such that an object at the normal distance (D) of the near point (i.e.25cm) from an image at a distance d cm away from the eye i.e., at the near point of hypermetropic eye.

PHYSICS

So u = + D cm and v = + d cm, and

1 1 1 1 1 = − = − f v u d D

Presbyopia: With advance of age, ciliary muscles become loose and the accommodation decreases. An appropriate concave lens is used to see clearly objects of near distance. But this lens does not work for the distant objects. Therefore, the upper half of the spectacle glass is made of one kind of lens and lower half of other kind. Astigmatism : This is a defect that rises when the cornea is not spherical but has larger curvature in one plane than in the other. This results in objects in one direction being well focussed while those in a perpendicular direction are not. This is corrected by glasses with lenses of cylindrical rather spherical shape. This defect is called astigmatism. --------------

PHYSICS NUMERICAL PROBLEMS REFLECTION 1. A concave mirror forms a real image four times in size of the object at a distance of 10 cm from the mirror. Find the radius of curvature of the mirror. 2. A convex mirror of radius of curvature 20 cm forms an image which is half the size of the object. Locate the position of the object. 3. A concave and convex mirror of radii of curvatures 15 and 25 cm respectively are placed 35 cm apart facing each other. An object of length 3 cm is placed in between them perpendicularly on their common axis at a distance of 10 cm from the concave mirror. Find the position, nature and size of the final image if the reflection first takes place in the concave mirror and then in the convex mirror. 4. A spherical mirror is placed against the wall of a room 25 feet long. An illuminated object is placed 11 ft in front of the mirror. What should be the type and focal length of the mirror in order that an image of the object may be formed on the opposite wall? Discuss the nature of the image in this case. 1 5. The image formed by a convex mirror is only of the size of the object. If the focal length of the 3 mirror is 12 cm, where is the object and where is the image? 6. An object is placed exactly midway between a concave mirror of radius of curvature 40 cm and a convex mirror of radius of curvature 30 cm. The mirrors face each other and are 50 cm apart. Determine the nature and position of the image formed by the successive reflections, first at the concave mirror and then at the convex mirror. 7. A plane mirror is placed 22.5 cm in front of a concave mirror of focal length 10 cm. Find where an object be placed between the two mirrors so that the first two images coincide. 8. A small object is placed on the principal axis of a concave spherical mirror of focal length 10 cm at a distance of 32 cm. By how much will the position and size of the image alter when a parallel sided slab of glass of thickness 6 cm and r.i. 1.5 is introduced between the centre of curvature and the object? The parallel sides of the blocks are perpendicular to the principal axis of the mirror. 9. An object is placed in front of a convex mirror at a distance of 50 cm. A plane mirror is introduced covering the lower half of the convex mirror. If the distance between the object and the plane mirror is 30 cm it is found that there is no parallax between the images formed by the two mirrors. What is the radius of curvature of the convex mirror? 10. A concave mirror forms on a screen, a real image of twice the linear dimension of the object. Object and screen are then moved until the image is three times the size of the object. If the shift of the screen is 25 cm determine the shift of the object and the focal length of the mirror. 11. An object is placed at a distance of 40cm from a convex spherical mirror of radius of curvature 20 cm. At what distance from the object should a plane mirror be placed so that the image in the spherical mirror and plane mirror may be in one plane?

12 Find the two positions of the object in front of a concave mirror of focal length 15 cm so that the image formed is three times the size of the object.

PHYSICS

13. A concave mirror of radius 40 cm lies on a horizontal table and water is filled in it up to a height of 5.0cm. A small dust particle floats on the water surface at a point P vertically above the point of contact of the mirror with the table. Locate the image of the dust particle as seen from a point directly above it. The refractive index of water is 1.33.

14. A source of light is at 10 cm from a convex mirror and is then moved up to a distance of 2cm from the mirror. How much does the image move if the radius of curvature of the mirror is 4.8cm? 15. Show by spherical mirror equation that in the case of a plane mirror, image is equal to object distance. 16. Prove that for the production of real image in a convex lens, the minimum distance between the object and the image is 4f, were f is the focal length of lens. 17. A convex lens of focal length 20 cm,(fig a) is cut into two equal parts so as to obtain two planoconvex lenses are shown in (fig b). The two parts are then put in contact as shown in (fig c). What is the focal length of combination?

18. An arrow of 15cm length is arranged along the axis of a concave mirror. The sharp edge is at 30cm from the mirror. If the radius of curvature of the mirror is 20 cm, then find the magnification of the arrow. REFRACTION 19. A point source of light S is placed at the bottom of a vessel containing a liquid of refractive index 5/3. A person is viewing the source from above the surface. There is an opaque disc of radius 1 cm floating on the surface. The centre of the disc lies vertically above the source. The liquid from the vessel is gradually drained out through a tap. What is the minimum height of the liquid for which the source cannot at all be seen from above? 20. There is a fish at a depth of 6 ft and at a distance 4.5 ft from the bank of a lake. A boy of height 5ft is standing at a distance 8ft from the side of the lake. How much the boy should be proceed such that the fish can visualize his shifting (movement)? The side of the lake may be taken as straight. Refractive index of water µ = 4/3. 21. A little water is poured in a concave mirror of curvature 30cm. A pin kept at a distance of 22cm from the mirror has its image coincident with itself. Find the refractive index of water. 22. A glass plate has a thickness t and index of refraction µ . The angle of incidence of a ray from air into the plate is equal to the critical angle for glass-air refraction. Calculate the displacement of the ray due to its passage through the plate.

PHYSICS

23. A slab of glass of refractive index 1.5 and thickness 3 cm is placed with the faces perpendicular to the principal axis of the concave mirror. If the radius of curvature of the mirror is 10cm, find where an object must be placed so that the image coincides with the object? 24. A ray of light falls on a transparent glass plate of refractive index 1.62 (w.r.t. air). If the reflected and refracted rays are mutually perpendicular, what is the angle of incidence? 25. A ray of light enters a rectangular glass slab of refractive index 3 at an angle of incidence 60°. It travels a distance of 5 cm inside the slab and emerges out of the slab. What is the perpendicular distance between the incident and the emergent ray? 26. The distance between two point sources of light is 24cm. Where should a convergent lens of focal length 9 cm be placed between them to obtain the images of both the sources at the same point. 27. A vessel of depth d is filled with a liquid of refractive index n1 and the other half is occupied by a liquid of refractive index n2 .Find the apparent depth of the vessel when viewed normally. 28. A ray of light is incident at an angle i on a glass slab on thickness t and the angle of refraction is r. t sin ( i − r ) . Show that the lateral displacement of the emergent ray is cos r 29. A parallel beam of light of width 1 cm is incident at an angle of 60° on a medium index of refraction 3 . Find the width of the refracted beam. 30. A small air bubble is situated inside a glass sphere of radius 2 cm. If the bubble is observed along a diameter, its apparent depth is 1cm. Calculate the actual depth of the bubble. 31. A spherical surface of radius R separates two media of refractive indices µ1 and µ 2 as shown in figure. Where should an object be placed in the medium 1 so that a real image is formed in medium 2 at the same distance?

32. A small object is placed at a distance of 20cm from one side of a glass block of thickness 10 cm. The opposite side of the block is silvered. The image is formed at a distance of 23.2 cm from the silvered face. Calculate the refractive index of glass. 33. A ray of light is incident on a glass plate of refractive index 1.62. If the reflected and refracted rays are mutually perpendicular, find the angle of incidence of the ray. PRISM 34. A prism is found to give a minimum deviation of 51°. The same prism gives a deviation of 62°48′ for two values of the angles of incidence, namely 40°6′ and 82°42′. Determine the refracting angle of the prism and the refractive index of the material. 35. Adjacent figure shows the path of a ray passing through an equiangular prism PQR. It is incident on face QR at the critical angle for total internal reflection. If the angle α shown in the figure is 40°, calculate the refractive index of the material of the prism.

PHYSICS

36. One face of a prism of refractive index 1.5 and angle 75° is covered with a liquid of refractive index 3 2 4 . What shouldd be the angle of incidence inciden of light on the clear face of the prism for which light is just totally reflected at the liquid covered face? 37. A ray of light undergoes deviation of 30° when incident on an equilateral prism of refractive index 2 . The angle made by ray inside the prism with the base of the prism is to be determined. 38. A glass prism with an apex angle of 60° has µ = 1.6. (a) What is the smallest angle of incidence for which a ray can enter one face of the prism and emerge emerge from the other? (b) What angle of incidence would be required for the ray to pass through the prism symmetrically? 39. Three right angled prisms of r.i. µ1 , µ 2 and µ 3 are placed side by side in such a manner that refracting angle of the second is just in opposite direction with respect to the refracting angle of 1st and 3rd one. If a light ray passed through this combination of prism without deviation show that µ12 + µ 32 − µ 2 2 = 1 . 40. A prism of material A deviates red light by 10° and blue light by an angle of 16°. Another prism of material B deviates red light by an angle 8° and blue light by an angle 14°. Which material have greater dispersive power? 41. i) Find the angle of incidence and also the angle angle of deviation of ray of light that passes symmetrically through a glass prism having refracting angle of 80°. 80 µ of glass = 1.5, sin 40° = 0.6428 , sin 74°37′ = 0.9642 (ii) What is the greatest value of the r.i. for which light can pass in this way through 80° prism? What is the corresponding angle of deviation? 42. A ray of light incident normally on one face of the faces of a right angled isosceles prism is found to be totally reflected. What is the minimum value of the refractive index of the material of the prism? When the prism is immersed in water, trace the path of the emergent rays for the same incident ray, indicating values of all the angles. ( µ u = 4 / 3 ) 43. A ray of light is incident at an angle of 60° on one face of a prism which has an angle of 30°. The ray emerging out of the prism makes an angle of 30° 30 with the incident ray. Show that the emergent ray is perpendicular to the face through throug which it emerges and calculate the refractive index of the material of the prism. 44. A ray of light is incident normally on one face of a prism and it totally gets reflected by other face. What happens to the ray if the prism is immersed in water? [R.I. of glass = 1.5, R.I of water = 1.33] 45. Light enters a prism of angle A at a grazing incident to emerge at an angle θ with the normal. Show that the refractive index of the material of the prism is given by.   cos A + sin θ 2  µ = 1 +    sin A    

1/ 2

PHYSICS

46. One of the refracting face of an isosceles prism is silvered. A ray of light, incident normally on the other refracting surface, emerges through the base of the prism normally after being reflected twice from the other two sides. Find the angles of the prism. 47. A glass prism of angle 72° and refractive index 1.66 is immersed in a liquid of refractive index 1.33. Find the angle of minimum deviation for a parallel beam of light passing through the prism. 48. Show that if the angle of prism is greater than twice the critical angle of its material, there would be no emergent ray whatever be the angle of incidence. THIN LENS 49. A plane-convex lens acts like a concave mirror of 28 cm focal length when its plane surface is silvered and like a concave mirror of 10 cm focal length when its curved surface is silvered. What is the r.i. of the material of the lens? 50. A converging lens of 20cm focal length is arranged coaxially with a diverging lens of focal length 8 cm. A point object lies on the same side as the converging lens and very far away on the axis. (i) What is the smallest possible distance between the lenses if the combination is to form a real image? (ii) If the lenses are placed 6 cm apart, what is the position and nature of the final image of the distant object? 51. Camera A having an f 8 lens, 2.5 cm in diameter, photographs an object using the correct exposure of 1/100 sec. What exposure should a camera B use in photographing the same object if it has an f 4 lens, 5 cm diameter? 52. Find the nature and focal length of a lens which must be placed in contact with a concave lens of focal length 25cm in order that the lens combination may produce a real image 5 times the size of the object placed 20 cm from the combination. 53. A concave mirror of focal length 20cm is placed at a distance of 25cm behind a concave lens. A pin is placed 68.6 cm in front of the lens forms an inverted image by the lens mirror combination which coincides with itself. With the help of a suitable ray diagram calculate the focal length of the lens. 54. The distance between an object and a screen in 96 cm. A convex lens is displaced in between them such that a clear image of the object is obtained on the screen for two positions of the lens. If the ratio of lengths of the images be 4.84, calculate the focal length of the lens. 55. The refractive indices for the substance of a lens for yellow and red colour of light are 1.54 and 1.51 respectively. The focal length of a lens for light of yellow colour is 30 cm, calculate the focal length of the lens for the light of red colour. 56. If a object is placed at a certain distance from a convex lens a real image is formed whose magnification is m1 and if the body is moved by a distance x, the magnification of the real image  1 1  − . becomes m2 . Prove that the focal length f = x   m2 m1  57. Light rays are falling on a convex lens as shown in figure. If the focal length of the lens is 30cm, then find the position of the image.

PHYSICS

58. An object is placed at a fixed distance from a screen. When a convex lens is placed between the object and the screen at a distance of 15 cm from the object, than an image is formed on the screen, having a size 5 times the size of the object. Calculate the focal length of the lens and the distance between the object and the screen. How much the lens be displaced towards the screen so that again a distinct image of the object be formed on the screen? What will be the magnification of this image? OPTICAL INSTRUMENT 59. A telescope has an object of focal length 50cm and an eyepiece of focal length 5cm. The least distance of distinct vision is 25cm. The telescope is focussed for distinct vision on a scale 200 cm away from the objective. Calculate (i) the separation between the objective and the eyepiece and (ii) the magnification produced? 60 The focal lengths of the objective and the eyepiece of a compound microscope are 1 cm and 5 cm respectively. An object, placed at a distance of 1.1cm from the objective has its final image formed at a distance of 25cm from the eye. Find the magnifying power of the microscope. 61. The objective and eyepiece of a microscope have focal lengths of 1 cm and 2 cm respectively and are separated by 12cm. A person whose distance of distinct vision is 25cm uses the microscope to see a small object. Where must the object be placed? 62. A compound microscope is composed of an objective and an eyepiece of focal length 0.5cm and 1.5cm respectively. Assuming that the least distance of distinct vision as 25cm, calculate the spacing required between the objective and eyepiece in order that the magnifying power is 500. 63. A person’s near point is 50cm and far point is 1.5m. What spectacles will be required (a) for reading purpose and (b) for seeing distant object? Least distance of distinct vision is 25cm. ANSWER 1. u = 10 cm, m = 4 Thus v = 4 × 10 = 40 cm (v is positive since the image is real) 1 1 2 1 1 2 40 Now, using + = we get, or, r = 2 × = 16 cm + = v u r 40 10 r 5 Since, this a concave mirror r is positive as expected. u 1 2 r = – 20 cm, m = thus v = 2 2 Since a convex mirror can form only virtual images the image distance is negative. 1 1 2 Using the equation + = v u r 1 1 2 2 1 1 1 1 or, or, − = − or, u = 10 cm We get, + = − + =− v u 20 u u 10 u 10

Therefore the object is at a distance of 10cm from the mirror. 3.

Here the image produced by the concave mirror acts as the object for the convex mirror. Now, for concave mirror, u = + 10 cm and r = + 15 cm 1 1 2 or, v = + 30 cm + = ∴ From mirror equation, v 10 15 Hence the image is formed 30 cm in front of the concave mirror i.e. image is real and inverted.

PHYSICS v 30 = = 3 ∴ Size of the image = 3 × 3 = 9 cm u 10 Now for convex mirror, u = 35 – 30 = 5 cm (+ve) and r = – 25 cm 1 1 2 25 or, v = − = −3.57 cm ∴ From mirror equation, + = − v 5 25 7 Hence the final image lies 3.57cm behind the convex mirror and it is virtual and erect w.r.t. the intermediate image i.e. the final image is inverted w.r.t. the object. 3.57 5 5 Also m2 = ∴ Size of the final image = × size of the intermediate image = 5 7 7 5 = × 9 = 6.43 cm 7 1 1 1 1 1 25 + 11 We have, u = 11ft; v = 25ft. ∴ = + = + = f v u 25 11 25 ×11

Also, m1 =

4.

5.

6.

∴ f = 7.64 ft. [Distance measured in the direction opposite to the incident ray is assumed positive]. As the focal length is +ve, so the mirror is concave. v 25 Also magnification, m = = = 2.27 u 11 Thus image is real, inverted and 2.27 times magnified than the object. 1 or, u = 3v m = v /u = 3 1 1 1 We know = + . Here u is negative. f v u 1 1 1 ∴ ∴ v = 8 cm behind the mirror ∴ u = 3v = 24 cm in front of the mirror. = − 12 v 3v The image formation is shown in the figure i) For concave mirror, u1 = 25 cm, f1 = 20 cm and v1 = ? Now,

1 1 1 = + f1 u1 v1

or,

1 1 1 = + 20 25 v1

∴ v1 = 100 cm As v1 is positive, hence the image is real. In the absence of convex mirror, the rays after reflection from concave mirror would have formed a real image I1 at distance 100 cm from the mirror. Due to presence of convex mirror, the rays are reflected and appear to come from I 2 . ii) In this case I1 acts as virtual object and I 2 is the virtual image. The distance of the virtual object from the convex mirror is 100 – 50 = 50 cm. Hence u2 = −50 cm. As focal length of convex mirror is negative and hence f 2 = − Here we shall calculate the value v2

30 = −15cm 2

PHYSICS

Using mirror formula, we have −

1 1 1 =− + 15 50 v2

or, v2 = −21.43cm

As v2 is negative, image is virtual. So image is formed behind the convex mirror at a distance of 7.

21.43 cm. In the adjoining figure F is the focus of the concave mirror M 1 . O is the position of the object. I is the position of the image due to reflection at plane mirror M 2 , which by question, is also the position of the image due to reflection at the concave mirror. By question, PQ = 22.5 cm. Let, PO = x cm. Then OQ = {22.5 – x} cm and QI = OQ ∴ PI = PQ + QI = 22.5 + ( 22.5 − x ) = {45 − x) cm

8.

9.

For reflection at the concave mirror, u = PO = x cm, v = PI = (45 – x) cm and f = 10 cm (given) 1 1 1 Therefore, from mirror equation, + = , solving we get x = 15 cm or, 30 cm 45 − x x 10 The distance between the two mirrors being 22.5 cm, the solution x = 30 cm is inadmissible, since in that case the object will be behind the plane mirror and no image of it can be formed due to reflection at plane mirror. Hence the object must be placed between the two mirrors 15 cm away from the concave one. P is the position of the object when there is no slab, u = 32 cm; f = 10 cm, v = ? 1 1 1 1 1 1 11 or, = − We have, + = = v 32 10 v 10 32 160 160 ∴ v= = 14.5 cm 11 v 160 ∴ Magnification, m = = = 0.45 u 11× 32 When the slab A is placed, the rays from P appear to come from P′, whose displacement from

 1 1   P = t  1 −  = 6 1 −  = 2cm  1.5   µ The distance P′ from O, the pole of the mirror, is therefore = 32 – 2 = 30cm 1 1 1 1 1 1 1 1 1 1 we get, + Again from, + = or, = − ∴ v = 15 cm = = v u f v 30 10 v 10 30 15 The image is displaced by (15 – 14.5) = 0.5 cm v 15 1 Present magnification = = = u 30 2 Let, O be the object placed in front of a convex mirror MM ′ at a distance of 50 cm as shown in figure. The distance of the plane mirror NN ′ from the object is 30 cm. We know that in a plane mirror the image is formed behind the mirror at the same distance as the object in front of it. It is also given that there is no parallax between the images formed by the two mirrors i.e., the image is formed at a distance of 30 cm behind the plane mirror. For convex mirror, u = 50 cm, v = – 10 cm as v is negative in convex mirror. (∵ QP = QN − PN )

PHYSICS

1 1 1 = + f u v 1 1 1 4 50 we have, = − =− ∴ f =− f 50 10 50 4 50 × 2 Now r = 2 f = − ∴ The radius of curvature of convex mirror is 25 cm. = − 25cm 4 1 1 1 1 1 1 v− f v v− f v v 10. We known = + or, = − = or, = = −1 ∴ m = −1 f v u u f v vf u f f f v v + 25 Now from the problem, we have, 2 = − 1 ... (1) and 3 = − 1 ... (2) f f Using the mirror formula,

Subtracting (1) from (2), 3 − 2 =

25 ∴ f = 25cm f

v − 1 ∴v = 75cm 25 1 1 1 1 1 1 3 −1 2 ∴ or, = + = − = = 75 u 25 u 25 75 75 75 Also from the second case, v′ = 75 + 25 = 100 cm

Again from (1), 2 =

∴ u = 37.5 cm

1 1 1 1 1 1 4 −1 3 100 or, or, u′ = cm + = = − = = 100 u′ 25 u′ 25 100 100 100 3 75 100 225 − 200 25 The shift of the object = − = = = 4.17 cm 2 3 6 6 1 1 1 11. For a convex mirror + = . Here u = – 40 cm, f = + 10 cm, v = ? v u f

∴

∴

1 1 1 + = v − 40 10

or, v = +8 cm

The distance between the object and its convex mirror image = 40 + 8 = 48 cm. Since in a plane mirror the image is as much behind the mirror as the object is in front of it, the 48 distance of the plane mirror from the object must be cm = 24 cm 2 12 Here two situations are possible. i) When real magnified image is formed v In this case u, v both are positive and f = 15. The magnification, m = = 3 or , v = 3u u 1 1 1 1 1 1 For a mirror = + or, = + . Solving we get u = 20 cm. f v u 15 u 3u ii) When virtual magnified image is formed. 1 1 1 Is this case v = −3u ∴ = − or, u = 10 cm 15 u 3u

PHYSICS

13. The image formation is shown in figure. First of all we shall consider the image formed by concave mirror. Consider the direction shown in figure. We have, R = – 40 cm and u = – 5 cm. Using the mirror formula, we get 1 1 2 + = u v R 1 2 1 or, = − v R u 1 2 1 6 or, = − = v −40 −5 40 40 ∴ v= cm = 6.67 cm 6 As v is positive, the image P1 is formed below the mirror (virtual). The reflected rays are refracted at water surface. The depth of point P1 from the water surface is 6.67 + 5.00 = 11.67. Due to presence of water, the image P1 will be shifted above by a distance 1   = 11.67 1 − = 2.92 cm  1.33  ∴ Final image is formed at a point (11.67 – 2.92) cm = 8.75 cm below water surface. 14. Focal length of the mirror f = R/2= 2.4 cm 1 1 1 Using mirror formula in first case, we have = + (u is negative) f v u 1 1 1 = + ∴ v = 1.94 cm behind the mirror v 10 2.4 1 1 1 In the second case : = or, v′ = 1.09 cm behind the mirror + v′ 2.4 2 The shift of image is (1.94 – 1.09) = 0.85 cm 1 1 2 15. From spherical mirror equation, + = . For a plane mirror, r = ∞ v u r 1 1 2 ∴ In the case of a plane mirror, + = = 0 ⇒ v = − u v u ∞ ∴ image distance = – (object distance) 16. Let, u and v be the object distance and image distance. Then the distance between the object and the image I = u + v dI d dv If I is minimum, = 0 or, ( u + v ) = 0 ⇒ = −1 du du du 1 1 1 For a real image, + = v u f

∴

1 1 1 = − 2.4 v 10

or,

1 dv 1 dv v2 Differentiating, − 2 − =0 ⇒ =− 2 v du u 2 du u 2 v So for I min − 2 = −1 ⇒ u = ±v u 1 1 1 Putting this condition in image equation, + = ⇒ 2 f = u u u f

PHYSICS ∴ I min = ( u + v ) min = 2 f + 2 f = 4 f (Proved)

17. A single double convex lens may be regarded as a combination of two plano-convex lenses placed in contact. Let f be the focal length of each plano-convex lens. 1 1 1 2 1 2 or, or, f = 40 cm Then = + = = F f f f 20 f Further, the two lenses of focal length 40 cm are put in contact. Therefore, the focal length of the combination again will be 20 cm. 18. If we consider the sharp edge of the arrow, then object distance u = 30cm, focal length f = 10 cm R  ∵ f = 2  1 1 1 + = ∴ From the relation, v u f 1 1 1 1 1 1 we have = − = − = ∴ v = 15 cm v f u 10 30 15 Again if we consider the last point of the arrow, then object distance will be (30 + 15) cm = 45cm 1 1 1 1 1 1 = − we have, [v1 = new image distance] = − ∴ From the relation, v f u v1 10 45 ∴ v1 = 12.85cm ∴ Length of the image = (15 – 12.85) cm = 2.15 cm ∴ Magnification of the arrow, m =

2.15 = 0.143 15

Refraction: 19. Suppose at a height h of the liquid in the beaker, an incident ray SK is incident at critical angle between water and air as shown in fig. Also any other ray will be totally internally reflected since for those rays the angle of incidence would be greater than critical angle. 1 3 GK 1 In ∆SGK , tan C = = . Also sin C = = GS h µ 5

9  4  ∴cos C = 1 −  =  25  5 3/5 3 1 3 Now tan C = = ∴ = or, h = 1.33cm 4/5 4 h 4 20. Let the point O at a depth of 6 ft from the surface of water (lake) indicates the position of the fish. Therefore OA = 6 ft and AB = 4.5 ft. PQ is the initial position of the boy at a distance 8 ft from the side of the lake. When the boy comes at RS towards the lake, his movements become visible to the fish. Let RB= x ft. From the figure µ sin i = sin r ... (i) Here µ = 4 3, sin i = AB

AB 2 + OA2 = 4.5

2

6 2 + ( 4.5 ) = 4.5

56.25 = 4.5 7.5 = 3 5

PHYSICS

and sin r =

BR 2

BR + RS

2

=

x 2

x + 25

4 3 × = x x 2 + 25 3 5 or, x = 20/3ft

From the equation (i), 16 x 2 + 400 = 25 x 2

or, 16 25 = x 2

(x

2

+ 25 )

20   Therefore, if the boy comes towards the lake by  8 −  = 4/3 ft = 1 ft 4 inch then his 3   movements become visible to the fish. 21. Let, P be the position of the pin. If the rays from P, after being refracted by water are incident on the spherical mirror at C normally, the rays will retrace their path and will from image at P. So, if CN is produced it will pass through the centre of curvature O of the concave mirror. In other words, CNO is a radius of curvature of the mirror. Let MN is drawn perpendicular to the surface of water at N. Here the angle of incidence i = ∠PNM = ∠NPA and the angle of refraction r = ∠CNE = ∠NOA sin i sin ∠NPA NA / PN ON If µ be the refractive index of water, then µ = = = = sin r sin ∠NOA NA / ON PN OA PA Again if the depth of water AB is negligible compared to PA or OA (amount of water is very OB small), then OA = OB and PA = PB . Hence µ = PB Now, OB = radius of curvature = 30 cm and PB =distance of the pin = 22cm ∴µ = 30 / 22 = 1.36

Since the ray PN is very near the principal axis of the mirror, ON = OA and PN = PA ∴µ =

22. Lateral displacement ( δ ) = RN = RQ sin ( C − r ) = t sec r sin ( C − r ) or, δ = t sec r ( sin C cos r − cos C sin r ) = t ( sin C − cos C tan r ) Now, sin C =

1 sin C 1 and µ = or, sin r = 2 sin r µ µ

1 1 1  t 1   ∴ δ = 1 −  ∴ δ = t  − 1− 2 . µ µ µ  µ 4 − 1  µ 2 + 1   23. Figure shows a concave mirror MM ′ and a glass slab AB perpendicular to principal axis. Here the problem is to find the position of object such that its image coincides with it. Let O be the required position. OBAN represents the path of the ray which retraces its path and forms the image at O. This is only possible when AN falls normally on the concave mirror. In another words we can say the if NA is produced it must pass through O′ i.e., the centre of curvature of the mirror. So O′ is the centre of curvature of concave mirror i.e., PO′ = 10 cm .

 1  1  The distance OO′ = t 1 −  = 3 1 −  = 1cm  µ   1.5 

PHYSICS

So the distance of the object O from the mirror = PO ′ + O′O = 10 + 1 = 11cm 24. A ray of light OA is incident on the glass plate and it is partly reflected (AB) and partly refracted (AC) at point A. In figure, ∠OAN = ∠NAB = i (Since the angle of incidence and angle of reflection are equal) Now, ∠N ′AC = r or, ∠i + ∠r = 90° or, r = 90° − 1 ∴ According to the question, ∠BAY + ∠CAY = 90° From Snell’s Law, a µ g =

sin i sin i = = tan i sin r sin ( 90° − 1)

or i = tan −1 (1.62 )

∴tan i = 1.62

25. Ray of light OA is incident on the block. At the point A it is refracted into the slab in the direction AB. At B the ray emerges out in direction BD which is parallel to incidence direction. Therefore, according to the question, AB = 5cm, i = 60°, µ = 3

BC = sin ( i − r ) or , BC = 5sin ( i − r ) AB sin i sin 60° But µ = or , 3 = sin r sin r

In ∆ABC ,

3

∴ sin r =

( 3) × 2

=

1 2

... (i)

∴ r = 30°

Therefore from equation (i) : BC = 5sin ( 60° − 30° ) = 5sin 30° = 2.5 cm Thus the perpendicular distance between the incident and emergent rays is 2.5 cm. 26. Evidently the convergent lens is so placed that the real image of one source coincides with the virtual image of the other source. Let u be the distance of the lens from the first source and its image is real. 1 1 1 1 1 1 ∴ − − =− ... (i) or, = − v u 9 v 9 u For virtual image of the second source, u1 = ( 24 − u ) and image distance = + v. 1 1 1 ∴ − =− v ( 24 − u ) 9

∴

1 1 1 = − v ( 24 − u ) 9

... (ii)

1 1 1 1 or, u 2 − 24u + 108 = 0 ∴ u = 6 or, 18 cm − = − 9 u 24 − u 9 So, the lens is to be placed at a distance of 18 cm from one source and 6 cm from the other source. 27. Apparent depth of the vessel = apparent depth (x) of the liquid of r.i. n1 + apparent depth ( v ) of

From equation (i) and (ii),

the liquid of r.i. n2 . Now, x =

Real depth of liquid of r.i. n1 d 2 = n1 n1

Again, v =

Real depth of liquid of r.i. n2 d 2 = n2 n2

PHYSICS

d d d1 1  + =  +  . This is the apparent depth of the vessel. 2n1 2n2 2  n1 n2  28. Lateral displacement, CE = BC sin ( i − r ) ∴ x+ y =

But BC = ∴ CE = t

BN t = cos r cos r sin ( i − r )

cos r 29. Let AB and CD denote the widths of incident and refracted beams respectively, AB = AC cos i , CD = AC cos r .

1 − ( sin i µ ) CD cos r 1 − sin 2 r = = = ∴ AB cos i cos i cos i

2

∴ CD = 3 cm = 1.732 cm 30. In the figure, A is the actual position of the bubble inside the sphere. When observed along the diameter OCD, B is the apparent position of the bubble. So, OB = v and OA = u and OC = r. µ µ µ − µ1 Now, for refraction at the he spherical surface, 2 − 1 = 2 v u r 1 1.5 1 − 1.5 1 1.5 1 5 ∴ − ∴ = =− = 1+ = 1 u 2 4 u 4 4 1.5 × 4 6 ∴ u= = = 1.2 5 5 31. Let, x be the distance of the object from the poles in medium 1. Here, we have u = – x, v = + x and R = + R.

 µ + µ1  x= 2 R  µ 2 − µ1  32. The apparent shift of the silvered face of the block ABCD towards the object O is, We know,

µ 2 µ1 µ 2 − µ1 − = v u R

Here,

µ 2 µ1 µ 2 − µ1 , − = +x −x +R

 µ −1   µ −1   1 y =t = 10  = 10 1 −     µ   µ   µ  1 10 ∴ u = x + t − y = 20 + 10 1 −  = 20 + µ  µ  1 10 or, 23.2 + 10 1 −  = 20 + ⇒ µ = 1.515 µ  µ Refractive index of glass = 1.515. 1.515 33. The incident, reflected and refracted rays are shown in the adjoining adjoin figure figure. It is obvious that, i + r = 90° or, r = 90° − i sin i sin i sin i Now, n = = = = tan i or, 1.62 = tan i sin r sin ( 90° − 1) cos i ∴ 23.2 + y = 20 +

10 µ

or, tan i = tan 58°19′ (from table of natural tangents)

∴ i = 58°19′

PHYSICS PRISM 34. The incident ncident ray is deviated through δ = 62°48′ when angle i = 40°6′ . From the principle of reversibility of light, it is clear from the adjacent figure igure that the emergent ray (for which angle r = 82°42′ ) is also deviated through the same angle δ . Now δ = (i + r ) − A or, A = ( i + r ) − δ = 40°6′ + 82°42′ − 62°48′ or, A = 60°, which hich is the refracting angle of the prism For minimum deviation i = r δ +A 51° + 60° or, i = min or, i = = 55°30′ , which Hence, δmin = 2i − A w is the angle of 2 2 incidence at minimum deviation The refractive index of the material of the prism is given by, by ( δ + A ) sin 51° + 60° sin min 2 2 µ= = = 1.648 A 60° sin sin 2 2 1 35. sin ic = , β = A − ic and µ = sin α / sin β. µ Now, sin β = sin ( A − ic ) = sin A cos ic − cos A sin ic But, sin ic

1/2 1 1 , therefore cos ic =   µ2 − 1 µ µ

(

)

1/2   1 1 Hence, sin α = µ sin β = µ ( sin A cos ic − cos A sin ic ) = µ sin A µ 2 − 1 − cos A µ µ  

(

)

1/ 2

2   sin α + cos A   which on simplification gives, µ = 1 +    sin A     Putting α = 40° and A = 60°, we get µ = 1.66 36. At the 2nd face AC the ray makes an angle of incidence r ′ such that i′ = 90°

∴ For refraction at the 2nd face AC, 1.5sin r ′ =

3 2 sin 90° 4

3 2 2 1 × = = sin 45° ∴ r ′ = 45° 4 3 2 But, A = r + r ′ = 75° ∴ r = 75° − 45° = 30° Or, sin r ′ =

∴ The required angle of incidence is given by sin i = 1.5sin r =

∴ i = sin −1 ( 3 / 4 )

3 × sin 30° = 3 / 4 2

PHYSICS δ   sin  A + m  2   37. We know that, µ = A sin 2

 60° + δm  sin  2   Let us calculate δm ∴ 2= sin 30° 1 1  60° + δ m  ∴ sin  = sin 45°  = 2 × sin 30° = 2 × = 2 2 2   ( 60° + δm ) = 45° ∴ δm = 30° Or, 2 So, 30° is the minimum deviation. In this case the refracted ray is parallel to the base. So, it makes zero angle with base. 1 1 38. a) We know that, sin C = = , C = 38°41′ µ 1.6

Now if the ray is to emerge from the other face the angle of incidence r2 at the second face must be less than C. r2 < 38°41′ ...(1)

Also r1 + r2 = A

or, r2 = 60° − r1

From equation (1) and (2), 60° − r1 < 38°41′ Now b)

sin i1 = 1.6 sin r1

or,

sin i1 = sin r1 1.6

or,

or, r1 > 21°19′ sin i1 > 0.3636 1.6

...(2) or, sin r1 > 0.3636 ∴ i1 > 35°35'

For symmetrical refraction i1 should be equal to i2 and r1 should be equal to r2 We know that, r1 + r2 = A Now

sin i1 = 1.6 sin r1

A = 30° 2

∴ r1 = r2 =

or, sin i1 = 1.6 × sin 30°

or, sin i1 = 0.8

∴ i1 = 53°8′

39. The arrangement in shown in the adjacent figure PQRST is the path of the light ray through the prism. The total deviation of the ray is zero. So, the incident ray PQ and the emergent ray from the point T are parallel. Let, at the point Q, the angle of incidence and the angle of refraction are i1 and r1 respectively. As the prism is right angled, so the angle of incidence at the point R = 90° − r1 [∵ r1 + r2 = 90°] . If r3 be the angle of refraction at the point R then the angle of incidence at the point S = 90° − r3 . If r4 be the angle of refraction at S then the angle of incidence at the point T = 90° − r4 . Again the emergent ray and the incident ray PQ are parallel to each other. So, the angle of emergence at the point T = 90° − i1 . Now at the point Q considering refraction, sin i1 = µ1 sin r1

or, sin 2 i1 = µ12 sin 2 r1

If we consider refraction at the point R then µ1 sin ( 90° − r1 ) = µ 2 sin r3 ∴ µ12 cos 2 r1 = µ 2 2 sin 2 r3

... (ii)

For the point S, µ 2 sin ( 90° − r3 ) = µ3 sin r4

∴ µ 2 2 cos 2 r3 = µ32 sin 2 r4 ... (iii)

... (i)

PHYSICS Ultimately for the point T, µ3 sin ( 90° − r4 ) = sin ( 90° − i1 ) Combining the equations (i), (ii), (iii) and (iv), µ12 + µ 32 = 1 + µ 22

∴ µ32 cos 2 r4 = cos 2 i1 ... (iv) ∴ µ12 + µ 32 − µ 22 = 1

40. In the prism A the angular dispersion of red and blue light, light δ A = 16 − 10 = 6° Similarly, in the prism B, angular dispersion of red and blue light, light δ B = 14 − 8 = 6° Clearly, δ A = δ B i.e., both the prism have the same angular dispersion. But in the prism A deviation of the intermediate ray = B=

16 + 10 = 13° and the same deviation for the prism 2

14 + 8 = 11° 2

6 6 , dispersive power of the prism B ( ωB ) = 13 11 ∴ω B > ω A , dispersive ispersive power of prism B > dispersive power of prism A.

∴ Dispersive power of prism A ( ω A ) =

41. i)

When light passes symmetrically through a prism the deviation is minimum. minimum A In that condition, r1 = r2 = and i1 = i2 . Now δ m = ( i1 + i2 ) − A = 2i1 − A . 2 A Also, r1 = = 80° / 2 = 40° 2 sin i1 If, i1 be the angle of incidence, incidence =µ sin r1 ∴ sin i1 = µ sin r1 = 1.5sin 40° = 1.5 × 0.6428 = 0.9642 ∴ sin i1 = sin 74°37′

or, i1 = 74°37′.

Again δ m = 2i1 − A = 2 × 74°37′ − 80 = 69°14′ A = 40° . Further, sin i1 = µ × sin 40° 2 Now, the value of sin i1 cannot be greater than 1, so we can write

ii) Since the ray passes symmetrically, r1 =

1 1 = = 1.56 sin 40° 0.6428 Here i1 = 90°, so, δ = 2i1 − A = 2 × 90° − 80° = 100°

( µ )max =

42. i)

Refer for fig (a) It is evident that ∠E = ∠F = 45° Since at B the ray AB is reflected, therefore ∠i = ∠r = 45° 1 1 For internal reflection, sin i ≥ sin c ≥ ≥ µ sin 45° ∴ µ min = 1.414

ii) When the prism is immersed in water, the path of the ray is shown in fig (b). ). As the incident ray is normal to the prism, it passes undeviated through water and glass. sin 45° sin 45° At point B, g µ w = or, sin r = ∴ ∠r = 48°35′ sin r g µw

PHYSICS

43. Given that, i1 = 60°,

A = 30°, δ = 30°

Now, δ = i1 + i2 − A ∴ 30° = 60° + i2 − 30°

or, i2 = 0

i.e., the ray emerges out perpendicular to the face through which it emerges. Now since i2 = 0, r2 = 0 From A = r1 + r2 ,

30° = r1 + 0

Hence, at first face, µ =

or, r1 = 30°

sin i1 sin 60° = = 3 sin r1 sin 30°

44. The ray PQ is incident normally on the face AB of the prism ABC. Then it is totally reflected from the face AC. So, r2 = θC when θC = critical angle between glass and water. ∴ r2 = sin −1 (1 / µ ) = sin −1 ( 2 / 3 ) = sin −1 ( 0.67 ) = 41.8°

Now, if the prism is immersed in water, refractive index of glass with respect to water is u µ g = 1.5 / 1.33 = 1.128 So, the critical angle at glass water interface is

θc′ = sin −1 (1/1.128 ) = sin −1 ( 0.886 ) = 62.46° ∴ r2 < θc′ So, if the prism is immersed in water then the rays will not be totally reflected. It will be refracted to water at the face AC. 45. As the incident ray grazes the refracting surface of the prism, so the angle of refraction equals to the critical angle. As A = r1 + r2 and here r1 = θc ∴ r2 = A − θc Now for refraction at the second surface, µ =

sin θ sin θ = sin r2 sin ( A − θc )

 µ2 − 1 1 ∴ sin θ = µ [sin A.cos θc − cos A.sin θc ] = µ sin A. − cos A.  µ µ    µ2 − 1  1 ∵ sin θc = , cos θc =  = µ 2 − 1sin A − cos A  µ µ  

Or,

( sin θ + cos A)

2

= ( µ 2 − 1) sin 2 A = µ 2 sin 2 A − sin 2 A

∴ µ sin A = sin A + ( sin θ + cos A ) 2

2

2

2

  sin θ + cos A  2  ∴ µ = 1 +    sin A    

46. From the figure, we have ∠PQK = 90° − A = 90° − ∠PQN ∠PQN = A = ∠NQR,

∠PQR = 2 A .

Again PQ RT

So, ∠PQR = ∠QRT = 2 A = ∠TRS ∴ ∠LRS = 90° − 2 A = 90° − θ ∴ θ = 2 A ∴ A + θ + 0 = 180° Or, A + 2 A + 2 A = 180° or, A = 36° and θ = 72°

1/2

PHYSICS

47. Here a µ g = 1.66,

∴ I µg =

a µg a µI

=

a

µ I = 1.33,

A = 72°

1.66 , now a µ g = 1.33

A + Dm 2 A sin 2

sin

or,

1.66 = 1.33

72° + Dm 2 72° sin 2

sin

∴ Dm = 22°24′

48. We have, A > 2C If i1 = 0°, r1 = 0°. But, r1 + r2 = A ∴ r2 = A If i = 90° , r1 = C

∴ r2 = A − C

∴ r2 > 2C

But, A > 2C

∴ r2 > C

Thus, whatever be the value of i1 ranging from 0° to 90° light is incident on the second face of the prism at an angle greater than critical angle and is totally reflected, so that there is no emergent ray. Thus, the limiting angle of a prism is twice the critical angle. THIN LENS 1 2 1 2 49. In the first case we have, = + = [∵ f m = ∞ ] ... (1) 28 f I f m f I In the second case we have,

1 2 1 = + 10 f I f m

... (2)

Where f m is the focal length of the curved silvered surface. R where R is the radius of curvature of the curved surface. 2 1 1 1 140 280 Subtracting (1) from (2) we get, = − or, cm = f m . R = 2 f m = cm f m 10 28 9 9

Hence, f m =

From (1) we have, f I = 28 × 2 = 56 cm . Now, Or µ − 1 = 50. i)

1 ( µ − 1) = fI R

R 280 = = 0.55 ∴µ = 1.55 f I 9 × 56

For the convex lens : since u1 = ∞ , therefore, the rays would converge towards its focus F1. But the concave lens diverges and produces the final image. For the concave lens : the point F1 would act as the virtual object for concave lens. u2 = − ( f1 − x ) = − ( 20 − x )

∴

− ( 20 − x ) + 8 8 ( 20 − x ) 1 1 1 1 1 = − =− + = ∴ v2 = v2 f 2 u2 8 20 − x 8 ( 20 − x ) x − 12

For the image to be real ( v2 ≥ 0 ) , the minimum value of x should be 12cm. 8 ( 20 − x ) 8 ( 20 − 6 ) 56 = = − cm (virtual) (from concave lens) x − 12 6 − 12 3 51. Focal length of the lens A is f A = 8 × 2.5 = 20cm . ii) Given, x = 6 cm ∴ v2 =

Focal length of the lens B is f B = 4 × 5 = 20cm  1  Now time of exposure ∝    f /n 

2

 1  1 ∴ Time of exposure for A is ∝  100  ( 20 / 8) 

2

PHYSICS 2

2  1  t 1  4 20  1 Time of exposure for B is t ∝  =  ×  = ∴t = second  ∴ 20 / 4 1/100 20 8 4 400 ( ) ( )     52. f 2 = – 25 cm, m = – 5, u = – 20 cm, f1 = ?

Now, m =

v u

or, − 5 =

v −20

or, v = +100 cm

1 1 1 1 1 1+ 5 6 100 50 = − = − = = ∴F = = cm F v u 100 −20 100 100 6 3 1 1 1 3 1 1 3 1 1 1 3+ 2 5 Also, = = or, = − or, + = or, = = F f1 f 2 50 f1 25 50 25 f1 f1 50 50

Again,

or, f1 = 10 cm

Hence, the lens is convex and its focal length is 10 cm. 53. Let f be the focal length. Since the image is formed at the same position, the rays have reflected their path and falls on the concave mirror MM ′ normally. Hence, PC = radius of curvature = 2 × 20 = 40 cm For the concave lens alone, NO = object distance = 68.6 cm NC = image distance = PC – NP = 40 – 25 = 15 cm 1 1 1 ∴ = − ∴ f = 19.2 cm f 15 68.6 54. Let the length of image I1 in first position of lens be y1 and that of I 2 in second position be y2 According to question,

y1 y2

= 4.84

or, y1 = 4.84 × y2

If v be the length of the object, then y =

( y1 × y2 ) =

4.84 × y2 × y2 = 2.2 y2 y2 v 1 = = or, u = 2.2v y u 2.2 96 or, v = = 30 cm 3.2

Linear magnification of lens = Length of image / Length of object = According to the question, u + v = 96

or, 2.2v + v = 96

∴ u = 96 − v = 96 − 30 = 66 cm 1 1 1 1 1 1 1 1 16 − = + = From the lens formula = − , we have = f v u f 30 −66 30 66 330 330 ∴f = = 20.625cm 16 55. If fY and f R are respective focal length and nY and nR are respective index of the lens for yellow and red light. Then

 1 1  1 = ( nY − 1)  −  fY  R1 R2 

1 1  1 = ( nR − 1)  −  fR  R1 R2 

From equations (1) and (2),

... (1) ... (2) f R nY − 1 = f Y nR − 1

or, f R =

nY − 1 × fY nR − 1

PHYSICS 0.54  1.54 − 1  ∴ fR =  × 30 = 31.76cm  × 30 = 0.51  1.51 − 1  56. For the convex lens ‘f’ is negative and as the image is real v is also negative. 1 1 1 1 1 1 1 u u u  v ∴ − − =− or, + = ∴ +1 = or, 1 + = ∵ m1 =  ... (i)  v u f v u f v f m1 f  u

Here fY = 30cm,

nY = 1.54, nR = 1.51

In the second case, object distance = u + x, therefore similarly, 1 + (ii) – (i) gives,

x 1 1 = − f m2 m1

∴ f =

1 u+x = m2 f

... (ii)

x (proved) 1/ m2 − 1/ m1

57.

The rays are appearing to meet at the point O. The point O will acts as a virtual object for the lens. Thus, v = CI , u = CO = +10 cm , focal length of the lens f = +30 cm 1 1 1 1 1 1 − = , we have − = v u f v 10 30 1 1 1 4 30 Or, = + = or, v = = +7.5cm v 10 30 30 4 v I v 58. Given the magnification of the image, m = = = 5 ∴ 5 = or, v = 75cm u O 15 ∴ The distance between the object and the screen is = v + u = 75 + 15 = 90 cm Hence using the formula

75 1 1 1  1   1 6 = − =  −−  = ∴ the focal length of the lens f = = 12.5cm 6 f v u  75   15  75 In the displaced position of the lens, the object and image distances are interchanged, therefore, the displacement of the lens x = v − u = 75 − 15 = 60 cm

The magnification of the image in second case is calculated as follows, 1 1 m1m2 = −1 or, m2 = − =− m1 5 OPTICAL INSTRUMENT 59. Given, f o = 50 cm, fe = 5 cm, D = least distance of distinct vision = 25cm

Distance of the object from the objective = u = 200 cm. ∴ Distance of the image from the objective = v , which is given by 1 1 1 1 1 1 1 1 1 200 + = or, + = or, = − ∴v = cm v u fo v 200 50 v 50 200 3 v 200 1 or, m1 = = u 3 × 200 3 Also final image formed by the eyepiece is at 25cm from the eyepiece and it is virtual.

∴ Magnification produced, m1 =

PHYSICS ∴ v′ = −25cm, ∴ u′ =

f e = 5 cm,

u′ = ? ∴

1 1 1 1 1 1 1 6 + = or, − + = or, = 25 u 5 v′ u ′ f e u ′ 25

v′ 6 25 and the magnification produced, m2 = = 25 × = 6 6 u′ 25

∴ Separation between the objective and the eyepiece = v + u ′ =

200 25 425 + = = 70.83cm and 3 6 6

1 total magnification = m1 × m2 = × 6 = 2 3

v0  D 1 +  , f0 = +1.0 cm, u0 = 1.1cm u0  fC  1 1 1 1 1 1 1 10 ∴ = − or, = − = + f 0 v0 u0 1 v0 ( −1.1) v0 11

60. M = −

∴

1 10 1 = 1− = v0 11 11

or, v0 = 11cm

∴ M =−

11  25   1 +  = − 60 1.1  5 

∴ M = − 60

61. For eye lens ve = – 25 cm, fe = 2 cm 1 27 50 1 1  = − +  = − or, ue = − cm ue 50 27  2 25  50 274 Hence distance of the real image from objective = 12 − = cm 27 27 274 For image formation by the objective, v0 = , f 0 = 1cm 27 1 1 1 27 1 1 27 247 274 Now or, or, u0 = − − = − = 1 or, = −1 = − = −1.11cm v0 u0 f 0 274 u0 u0 274 274 247 1 1 1 − = ve ue f e

∴

1 1 1 − = −25 ue 2

or,

62. Given, f 0 = 0.5cm, fe = 1.5cm, m = 500. Let I be the spacing between the lenses. Now distance of the image formed by the eyepiece from it = v′ = −25cm (virtual) 1 1  1   1  1 1 For image formed by the eyepiece,   +   =   or, −   +   =  v′   u ′   f e   25   u ′  1.5  75  Distance of the image formed by the objective from it is v = I −   cm  53  From image formed by the objective for the object at u from the objective, 1 1  1   +  =   v   u   f0 

1 1 1 or,   +   =  v   u  0.5

or,

∴ u′ =

75 cm 53

1  1  ( 2v − 1) = 2−  = u v v

m1 = magnification by the objective =

 ( 2v − 1)  v   75    203  = v×  = 2v − 1 = 2  I −    − 1 = 2 I −   u  53    53    v 

m2 = magnification by the eyepiece =

v′ 53 53 = 25 × = u′ 75 3

  203   53 m = total magnification = m1 × m2 = 2I −    × = 500 (given)  53   3  (106 I − 203) = 500 ∴ I = (1500 + 203) = 16.06 cm Or, 3 106

PHYSICS

63. The person suffers from myopia as well as hypermetropia. a)

For correcting hypermetropia, the focal length required is :

1 1 1 1 1 = − = − f D d 25 50

100 100 = = +2 D f 50 For correcting myopia the focal length required is : f = − d = −1.5 m

Or, P = b)

Power of the lens = −

1 = − 0.667D 1.5 SHORT QUESTIONS

REFLECTION 1. If a mirror is rotated through an angle θ , the reflected ray rotates through 2θ . Is it always true? 2. Why are the multiple images formed by a thick mirror? Which image is brightest? 3. Which paper is better for reading - glossy or dull? 4. A mirror can reverse right or left, can it reverse up and down? 5. Why do the polished shoes look shinning? 6. Outer views are not visible from a strongly illuminated room surrounded by glass at night. But when we put out the lamps inside the room, outside becomes visible. Explain the cause. 7. The scale of a moving coil galvanometer with lamp and scale arrangement is at a distance 2m from the mirror. When d.c. is supplied to the coil, light point shows a deflection of 40 cm. What is the angular deflection of the coil? 8. Why does the colour of the moist objects seen richer and deeper than those of the dry objects? 9. A real image of an object is formed when light rays from the object incident on a concave mirror. If both of the object and mirror are immersed in water does any change in the position of the image occur? 10. What will be the magnification when an object is placed in between the pole and focus of a concave mirror? 11. What types of reflector are used in search light or headlight of a motor car? What is the reason? 12. A plane mirror is fixed at a height ‘h’ above the bottom of a beaker containing water (refractive index µ ) up to a height d. What would be the position of the bottom formed by the mirror? REFRACTION 13. A monochromatic beam of light is refracted from vacuum into a medium of refractive index ‘n’. What is relation between wavelength of incident and refracted waves? 14. Can the wavelength of light be determined using reflection or refraction phenomenon?

15. Refractive index of medium with respect to air is 2 and a ray from air makes an incident of 45° (incident angle 45°) in this medium. Find the deviation of the ray due to refraction. 16. A glass rod becomes invisible when immersed in glycerine-why? 17. If refractive index of diamond is 2.42, prove that all the rays of incident angle greater than 25° will reflected totally? 18. Why does a diamond sparkle more than a glass imitation cut to the same size? 19. Is it possible for the absolute refractive index of a medium to be less than 1? 20. Why are danger signal red while the eye is more sensitive to yellow? 21. What are conjugate points in case of a spherical mirror? If a pair of conjugate points be at the same side of a mirror. What type of mirror will it be?

PHYSICS

22. A street light, viewed by reflection across a body of water in which there are ripples, appears very elongated. Explain. 23. Can light be ‘piped’ like sound in a doctor’s stethoscope? 24. Why does the setting or rising sun appear to be oval? PRISM 25. A beam of white light passing through a hollow prism gives no spectrum. Explain. 26. Which of blue and red lights will be deviated more by a prism and why? 27. Why right angled isosceles prisms are used for reflection in periscope instead of plane mirrors? 28. If a prism is surrounded by a medium denser than the material of the prism, in which direction will a ray of light be deviated when it is refracted through the prism? 29. When a green body is illuminated by white and red light alternatively. What will be the colour of the body (appears)? Answer the explanation. 30. How does the dispersion of light takes place in vacuum? 31. Why is a rainbow formed in the sky? THIN LENS 32. Does the focal length of a thin convex lens increases or remain the same if monochromatic red light is used instead of blue light to measure it? 33. What is zoom lens? 34. Are all convex lenses converging? 35. What will be the nature of the lens formed by an air bubble in water? 36. A swimmer sees only hazy contours of object when he opens his eyes under water while they are distinct when using a mask, why? 37. Does the image from at infinity if an object is placed at the focus of a concave lens? 38. Is it possible for a given lens to act as a converging lens in one medium and as a diverging lens in another? 39. What do you mean by a chromatic combination? 40. A convergent beam is placed on a screen. How the point of convergence shifted it we place a parallel glass plate before the screen. Draw the ray diagram. OPTICAL INSTRUMENT 41. What is astigmatism? 42. What is resolving power of optical instrument? 43. How is astigmatism corrected? 44. What is the cause of presbyopia? 45. Is human eye equally sensitive for light of all wavelengths (colours)? 46. Why is the diameter of the objective of an astronomical telescope made large? 47. In what type of telescope is the final image erect? 48. What is colour blindness? 49. Reading strains the eyes more than viewing distant objects. Explain. 50. It is difficult to thread a needle with one eye closed. Why? ANSWERS REFLECTION 1. No, it is not always true. If the mirror rotates about the normal at the point of incidence, no rotation of the reflected ray occurs.

PHYSICS

2.

3.

4. 5.

6.

7.

Multiple images are formed by the thick mirror due to reflections from the front surface as well as from the silvered back surface of the mirror. The second image is the brightest as it is formed by the refracted part of the incident beam. Dull paper is a diffusely reflecting surface. A diffuse reflecting surface can be seen from any point in front of the surface, this is independent of the direction along which the incident light travels. The surface appears equally bright from all directions. That is why dull paper is better for reading than glossy paper. The incident and reflected rays lie in the same plane. So the image cannot reverse up and down. The surface of the leather is usually rough and from the shoes made of leather diffused reflection of light takes place and hence shoes do not look shinning. On polishing the surface of the leather becomes smooth and due to regular reflection of light from the polished surface the shoes look shinning. Light from outer bodies of the surroundings enter into the room. At the same time, strong light inside the room after reflection from glass reaches the observer. But intensity of reflected light is comparatively very high, so the bodies cannot be seen distinctly. But when the light inside the room is switched off, only light from outer objects reach the observer and hence the observer can see the outer objects easily.

M1M1′ is the initial position of the mirror of the galvanometer. Now when the point (optical point) goes from O to O1 , the deflection of mirror is θ . The final position of the mirror is M 2 M 2′ . So, the angle between the

reflected rays = 2θ. Now tan 2θ = 40 / 200 ∴ 2θ = 1/ 5 radian ∴ θ = 1 /10 and = 5.72° ∴ The angular deflection of the coil = 5.72° 8. A thin film of water covering a moist object reflects the incident white light in a particular direction and its own colour becomes predominant. Dry objects diffuse white light in all directions. That is why the colours of the moist objects are seen richer and deeper than those at the dry objects. 9. In case of concave mirror, if u be the object distance, v be the image distance and f be the focal 1 1 1 length, then the relation between u, v and f is + = . From this relation we can evaluate the v u f image distance for a definite object distance. Since u and v are independent of medium, so the above relation is also independent of medium. So, if the object and mirror are immersed in water, no change in image distance v will take place. f 1 1 1 1 1 2 1 10. Here image is virtual, u = Now, = − or, = − = − 2 v f u v f f f v f ∴ Magnification, m = = =2 ∴ v=−f u f /2 11. Concave mirror is used as reflector is search light as head light of a motor-car. When a light source is placed at the centre of curvature of concave mirror, the light rays of back-side after reflection (by the mirror) become parallel and traverse in forward direction. So, the amount of light in the front-side increases by a large amount. So, it becomes easier to visualize the objects at night.

PHYSICS

 1 12. The bottom of the beaker appears to be shifted up by a distance x = 1 −  d . the apparent  µ  1 d distance of the bottom from the mirror = h − 1 −  d = h − d + . The image will be formed µ  µ d behind the mirror at a distance h − d + µ REFRACTION Velocity of light in vacuum 13. R.I. n = Velocity of light in medium If λ and λ′ are the wavelength of light at vacuum and in medium respectively and γ be the frequency of light waves. Then n =

γ × λ λ Wavelength in vacuum = = γ × λ′ λ′ Wavelength in medium

14. Although the ratio of two wavelengths can be determined by refraction, absolute value of wavelength cannot be determined using reflection and refraction phenomenon. 15. The incident is shown in the adjacent ray diagram. If i be the angle of incidence and r be the angle of refraction, then from Snell’s Law, sin i sin i sin 45° 1 µ= ∴ sin r = = = ∴ r = 30° sin r µ 2 2 ∴ The required angle of deviation δ = i − r = 45° − 30° = 15° 16. We know that when an object of same R.I (as the medium) is placed in a medium, the object becomes invisible. Because as the refractive index is same, no reflection or refraction occurs from the plane of separation of the media. So, the object cannot be seen. Now the R.I.’s of glass and glycerine are equal. That is why when a glass rod is immersed in glycerine, it becomes invisible. 17. If θc be the critical angle when refraction occurs from diamond to air, sin θc =1 µ =1 2.42 = 0.4132 . ∴ θc = 24.4° . But we know that if the angle of incidence of light is greater than the critical angle of the medium, total reflection of the light occurs. As the critical angle of diamond is 24.4° , all of the rays whose angle of incidence are greater than 25°, get totally reflected. 18. The critical angle of glass is about 40.5° where as the critical angle for diamond is 24.5°. Thus a larger number of multiple reflections occurs inside diamond than in glass. Light entering from different faces go through a number of reflections and hence produce sparkling effect. Velocity of light in vacuum (c) 19. The absolute refractive index of a medium = Velocity of light in medium (v) The velocity of light is the highest in vacuum. For all other media v < c or, c v > 1 . So, the absolute refractive index of any medium will always be greater than 1. −4

20. Scattering is proportional to ( wave length ) . Now wavelength of red colour is more than the wavelength of yellow colour. So red light is scattered less than yellow light. So range of visibility is more for red light than yellow light. That’s why red light is used in danger signal. 21. The points on the principal axis which are such that if an object is placed at one point image will be obtained at the other are called as the conjugate points. The mirror is a concave one.

PHYSICS

22. Ripples which are large in number form images almost in continuation of each other and hence the impression of a very elongated light. 23. Yes, light can be piped from one point to another with little loss by allowing it to enter one end of a rod of transparent material. The light will undergo total internal reflection at the wall of the rod and will proceed along its axis. This has actually been achieved by using a bundle of glass fibres. 24. This is due to the unequal refraction of rays diverging from the lower and upper ends of the Sun. Rays from the lower edge have to go through a greater thickness of air than rays from the upper edge. So the vertical diameter appears to be diminished in size, whereas the horizontal diameter remains unaltered. PRISM 25. The situation is shown in figure. The faces AB and AC of the hollow prism act like plates. PQ is a ray of white light incident on face AB. This is refracted along QR and emerges as RS in air parallel to PQ. Now it falls on the face AC, refracted along ST and finally emerges along TV. As the rays of all colours from hollow prism in the same direction, there is no dispersion and so no spectrum is obtained. 26. Blue light will be deviated more than red light. Deviation δ = ( µ − 1) A , where µ is the refractive index of prism material and A is the angle of prism. The refractive index increases with decreasing wavelength. Since λ Red > λ Blue , µ B > µ R i.e., refractive index in case of blue light is greater than the refractive index of red light. Since the amount of deviation increases with the refractive index of the material of the prism, therefore, it can be said that deviation of blue light will be more than the deviation of red light. 27. Total internal reflection takes place in right-angled isosceles prisms used for reflection in periscopes to produce bright image. Moreover, this can be used for a long time as mercury layer is not given for reflection unlike plane mirrors. 28 When refractive index of material of the prism is smaller than that of the surrounding medium, the emergent ray bends towards the edge of the prism. 29. When the green object is illuminated by white light, the object looks green. Since, white light is a combination of green and other six colours, when white light falls on the object, the green coloured light is reflected by the object and rest of the colours are absorbed. So, the object looks green. But if the object is illuminated by red light, it looks black because now the green object absorbs red light as we know that absence of all colours is the reason of black colour. 30. The separation of colours present in a polychromatic light is called dispersion. Lights of different colours travel with different velocities in a material medium. This gives rise the phenomenon of dispersion. But in vacuum all the seven colours of light travel with same velocity. Hence, no dispersion of light is possible in vacuum. 31. Rainbow is produced if the light rays incident on the rain drops undergo refraction, dispersion and total internal reflection in order. THIN LENS 32. R.I. µ of the material of the lens depends on the colour of the light used. The value of µ is higher for blue colour than for red colour. Now focal length f of a lens is

1 1 1 = ( µ − 1)  −  where r1 and r2 the radii of curvature of the lens f  r1 r2 

surface are and µ is the refractive index of the material of the lens.

PHYSICS Here µ Blue > µ Red

33. 34. 35. 36.

37.

38.

39.

40.

∴ f Red > f Blue

Thus we find that focal length of a lens would be larger in the case of monochromatic light than that of blue light. Zoom lens is a combination of lenses with variable focal length. All convex lenses are not connverging. ing. If the refractive index of the surrounding medium be more than the refractive ive index of the material of the lens, then the lens would behave as diverging lens. The air bubble would act as a diverging lens because the refractive index of the surrounding medium (water) is greater than that of the material (air) of the lens. Inside water, the focal length of the eye lens increases, since R.I. .I. of water is larger than that of air.

  100 Consequently the power of the eye decreases  power =  .Therefore the image of the focal length   immersed body cannot be formed at retina. So the swimmer immer would see only lazy contours of the objects. But in case of masks being used, light rays enter the eye from air and that the focal length as also the power of the eye does not alter. The image will not form at infinity but will form on principal axis in middle point between the 1 1 1 optical-centre centre and focus. Because we know that − = . In this equation putting u = f we get v v u f = f/2. So the image will form at the middle point between the optical-centre optical centre and focus. Yes, when a glasss of biconvex lens is immersed in a liquid of refractive index greater than the refractive index of glass. It will behave like a divergent lens. In air, the same lens behaves as a convergent lens. Combination of two or more lenses, made of different materials materials with different dispersion power used to overcome chromatic aberration is called achromatic combination. The lens combination has the same desired focal length for two or more wavelengths. A glass slab is placed before the point of convergence and the ray will meet away from the previous convergence point which is shown in figure. O is the th point of previous convergence. O1 is the present point of convergence.

OPTICAL INSTRUMENT 41. The front surface of the cornea loses its its spherical shape and its curvature becomes unequal in different meridian planes. This defect is known as astigmatism. The focal length of the lens system of an astigmatic eye becomes different in different meridian planes. As a consequence, a line object (or lines of the same object) placed at the same distance but differently oriented cannot be seen by the eye equally distinct as their images are not clearly focused on the retina simultaneously. 42. The ability of an optical instrument to enable us to see two close objects distinctly is called the resolving power of the instrument. Two close points are said to be resolved, if they can be seen as separate. The closer these two distinguishable point objects are, the higher is the resolving power. 43. The defect ect is removed by using a cylindrical lens which has different curvatures in different meridian planes. The curvature and the axis of the lens are so chosen that together with the eye, the cylindrical lens forms an optical system whose focal length is the same in every meridian plane.

PHYSICS

44. Generally with age a normal eye gradually loses its power of accommodation and the near point recedes away from the eye than the normal distance of distinct vision i.e. 25 cm, such an eye is called presbyopic or farsighted eye and the defect is termed by the physician as presbyopia. 45. Human eye is not equally sensitive to light of all the colours. It is much more sensitive to light of yellow and green colour. 46. The diameter of the objective of an astronomical telescope is made large to collect more light from the object and to increase the brightness of the images. 47. In terrestrial telescope final image is erect. 48. Colour blindness is a defect of vision in which a person cannot distinguish between colours. This is due to damage or insensitivity of one, two or all the three sets of nerve which are sensitive to the primary colours, namely red, green and blue. 49. In the eye, the distance between the lens and the retina is fixed. Objects at different distances from the eye are focused on to the retina by altering the focal length of the lens (this is called accommodation). While viewing distant objects, the lens acquires a longer focal length, and is, therefore, less curved and thin. For reading, the lens must have a shorter focal length. It becomes more curved and thick. This causes the eye to be strained. 50. Because it is difficult to estimate the distance of the object with one eye. With both eyes open, the distance can be correctly estimated and threading the needle becomes easier. ----------