Unit 3 Problems Solutions

CC2404 Applied Physics and Instrumentation in Health Care Problems Unit 3 1. What is the power of the eye when viewing an object 50.0 cm away? (Take the lens-to-retina distance to be 2.00 cm) P=

1 1 1 1 1 = + = + = 52 .0 D f u v 0.5m 0.02 m

2. Calculate the power of the eye when viewing an object 3.00 m away. (Take the lens-to-retina distance to be 2.00 cm) P=

1 1 1 1 1 = + = + = 50 .3D f u v 3.0m 0.02 m

3. You note that your prescription for new eyeglasses is −4.5 D. What is the focal length? f =

1 = −0.222 m (concave lens) P

4. What is the size of the image on the retina of a 1.20× 10-2 cm diameter human hair, held at arm’s length (60.0 cm) away? (Take the lens-to-retina distance to be 2.00 cm) y' v v 2.00 cm = ⇒ y ' = y × =1.20 ×10 −2 cm × = 4 ×10 −4 cm y u u 60 .0cm

5. Calculate the power of the eye when viewing objects at the greatest and smallest distances possible with normal vision, assuming a lens-to-retina distance of 2.00 cm. 1

1

1

1

1

1

1

greatest distance: P = f = u + v = ∞ + 0.02 m = 50 .0 D 1

1

1

smallest distance: P = f = u + v = 0.25 m + 0.02 m = 54 .0 D 6. Optical fibers are used with endoscope to examine the inner structures of the body, which physical phenomenon is basis for image transmission in optical fibers? What are the other functions of an endoscope? Total internal reflection is the basis for image transmission in optical fibers. Perform minor surgical procedures; obtain samples, burn away obstructing plaque in major arteries

7. An eye has a near point of 15 cm and a far point of 50 cm. The distance of the retina from the front surface of the cornea is 0.019 m. For all calculations assume that the eye acts as a single thin lens situated at the front surface of the cornea. a. Calculate the power of the eye when focusing on an object at the far point of 50 cm. P=

1 1 1 1 1 = + = + = 54 .6 D f u v 0.5 0.019

b. Calculate the power that this eye should have when focusing on an object at infinity. P=

1 1 1 1 1 = + = + = 52 .6 D f u v ∞ 0.019

c. Calculate the power of the corrective lens needed by this eye to focus clearly on an object at infinity. 52 .6 D − 54 .6 D = −2.0 D d. State the eye defect and describe the shape of the corrective lens. short sighted (myopia) ; concave lens e. Calculate the power of the unaided eye when viewing an object at its near point. P=

1 1 1 1 1 = + = + = 59 .3D f u v 0.15 0.019

f. With the corrective lens in place, calculate the new position of the near point for this eye. 59 .3D − 2.0 D = 57 .3D P=

1 1 + u v

57 .3D =

1 1 + u 0.019

u = 0.21m

8. a. What power of contact lens is needed to correct the vision of a person who has a far point of 1.00 m? P=

1 1 1 1 1 = + = + = 51 .0 D f u v 1.0m 0.02 m

since the power of the normal eye viewing an object at the greatest distance is 50.0 D. the power of the contact lens = 50.0 D − 51.0 D =− 1.0 D b. If this person has a normal ability to accommodate (increase) the power of her eyes by 8.00%, what is the closest object she will be able to see clearly with contacts lens on? The power of her eyes for the nearest object: 51.0 D × 108% = 55.08 D With the contact lens on, the new power is: 55.08 D – 1.0 OD = 54.08 D P=

1 1 1 1 1 = + = + = 54 .08 D f u v u 0.02 m

u = 0.245 m