Question

1. Mr. Magoo is in desperate need of corrective lenses for his 3 cm long nearsighted eyeballs. He wants to see Comet Lulin and the stars tonight, but his far point is only 4 cm! (His near point is just 2 cm!). a. What type of eyeglass lenses and strength in diopters will you prescribe for him, if his eyeglasses will be 1.5 cm from his eye? b. Will his new glasses create a real or virtual image? Explain. c. Testing his new eyewear, Mr. Magoo tries to see an object 6 cm from his eye. Will the image be visible to him? Calculate both image distances and focal lengths. (Hint: use the information about his far point to find the focal length of his eye and treat this as a fixed value) d. Use ray tracing to locate the final image for the object at 6 cm using only the principal rays. Glasscy Exclens 14 15 I 2 3 4 5 6 7 8 9 10 11 12 13 e. Is the final image real or virtual? Inverted or upright? f. Should you show Mr. Magoo his bill before or after you fit him with new glasses?

can u please type your answer, thanks in advanced

Answer 1

a) Correction of nearsightedness requires a diverging lens that compensates for the overconvergence by the eye. The diverging lens produces an image closer to the eye than the object, so that the nearsighted person can see it clearly. Mr. Magoo who is nearsighted person should be able to see very distant objects clearly. That means the spectacle lens must produce an image 4.0 cm from the eye for an object very far away since 4 cm is his far point. An image 4 cm from the eye will be 2.5 cm from the spectacle lens since eyeglasses are 1.5cm from eye, hence 4-1.5 = 2.5cm. Therefore, we must get image distance di = −2.5 cm when Object distance is do ≈ ∞. The image distance is negative, because it is on the same side of the spectacle as the object.

Power of the glasses be P.

P=1/do+1/di         ----(1)

P=1/∞+1/−0.025 m Since 1/∞=0 , we obtain: P=0−40m=−40 D

Power is negative indicating concave lens, i.e.diverging lens of power -40D will be recommended for Mr. Magoo.

b) diverging lens glasses will create a virtual image since it is a diverginging lens and so rays seem to diverge from the lens's center and it will produce a virtual image of far away object near to the eye.

c) Object is 6cm from eye and glass is 1.5cm from eye, hence object distance do is 6-1.5 = 4.5cm. Power of lens is -40D.

Using the equation (1),

P = 1/do + 1/di

-40 = 1/0.045 + 1/di

1/di = -40 -1/0.045 = -62.22

di = -0.016m = -1.6cm.

Hence Mr. Magoo won't be able to see the object since the focal length is 1/P = 1/-40 = -0.025m = -2.5cm.

d)

eyelin Glasses image formed at P Ray of is back extended A P image is oximately appre at 1.5cm edia same side of object eve 14.5cm 2.5cm 6cm 2

The two principal rays are :

one passing through the center of lens undeviated and other is parallel to optic axis and then back extended to pass through focal point of lens(calculated as 1/P and P is calculated in a)). The two rays intersect at point p, hence image is formed at p. The image distance from glasses is approximately 1.5cm, by calculations it is 1.6cm. Hence there are ray sketching inefficiency and you may correct it while drawing on the actual sheet given and get exactly 1.6cm.

e) the image is virtual and upright.

f) the bill should be shown before as after fixing the glasses, he won't be able to see the bill which may be approximately at 6cm.

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