Question

Part II: Corrective Eyeglasses Not everybody can adjust their eyes to be able to see both distant and nearby objects clearly. In a normal eye, light from a distant object forms a sharp image on the retina. In a far-sighted eye, light from a distant object converges too gradually and would form a sharp image behind the retina (of course the retina gets in the way, and thus a blurry image hits the retina). In a near-sighted eye light from a distant object converges too quickly and would form a sharp image in front of the retina for nothing is there to detect the light, so the rays cross and begin to spread out, and thus a blurry image hits the retina) We will simulate this using the optical bench lamp with its letter F as our distant object. We will use an f= +10 cm lens to represent the eye lens. And a screen will serve as the retina. However, unlike the eye that has the ability to change its focal length, we are using rigid glass lenses, so we will simulate the eye defects by repositioning the lens. Concentrate on whether the image forms in front of, directly on, or behind the screen/retina. • Set the optical bench lamp and F at one end of the optical bench. • Place the screen at a distance of 80 cm from the F object. front of screen= near sighted back of screen= far sighted on screen=normal 237 Physics 8B Lab 7 - Optical Instruments rev 4.0 a) Place an f = +10 cm lens in position to represent a normal eye. Draw a ray- tracing diagram for this arrangement. b) Move the f = +10 cm lens to a position to represent a far-sighted eye. Draw a ray-tracing diagram for this arrangement

Answer 1

given, focal length of the lens used is f = +10 cm
distance between screen and object, D = 80 cm

a. For the normal eye
1/v - 1/u = 1/f
also, v - u = 80 cm
u = v - 80
f = vu/(u - v) = (80v - v^2)/(80)
v^2 - 80v + 800 = 0
v = (80 +- sqrt(6400 - 3200))/2 = 68.28427 cm , 11.71572 cm
now, eye lens is closer to retina than object, so similiarly we chose the configuration where lens is closer to screen
hence, v =11.72 cm
u = 68.28 cm
hence, lens is located 11.72 cm from the screen
the ray tracing diagram will be as under

Normal Eyel 80 cm

b. consider a lens is moved a bit to left, 12 cm from the screen
then
u = -78 cm
1/v - 1/u = 1/f
1/v = 1/10 - 1/78
v = 11.4705882
so image has moved ahead of the screen
hence, for far sighted eye, the lens has to be moved closer to the screen for the image to be formed behind the lens, the ray diagram is as under

---------- Far signted ness 80 cm

c. For a farsighted eye, for the image to converge a bit early, converging lens has to be added to the configuration touching the outher lens to converge the light to the screen
the ray diagram is as under

Near signtedness 80 cm

d. For near sighted eye, the lens moves away from the screen
the ray diagram is as under

Corrected farsigntedness by using a second converging lens 80 cm

e. for near sighted a diverging lens is used to divergethe rays a bit more and hence ensuring image is formed at the screen and not behind the screen, the ray diagram is as under

Far sightedness corrected with adding • another diverging lens: 80 cm

f. The image for a normal eye is real, and hence inverted, and corrective lens does not change the overall orientation of the final image, otherwise one would see the world upside down which is actually not the case for optical users