1)Two identical diverging lenses are separated by 17 cm. The focal length of eac

Question

1)Two identical diverging lenses are separated by 17 cm. The focal length of each lens is -8.0 cm. An object is located 4.0 cm to the left of the lens that is on the left. Determine the final image distance relative to the lens on the right.

*** its not -5.13 or 5.13 CM

2)A farsighted man uses eyeglasses with a refractive power of 2.10 diopters. Wearing the glasses 0.03 m from his eyes, he is able to read books held no closer than 0.28 m from his eyes. He would like a prescription for contact lenses to serve the same purpose. What is the correct contact lens prescription, in diopters?

** its not +1.86 diopters

Explanation / Answer

a) 1/f = 1/u + 1/v

1/-8 = 1/4 + 1/v

v = -2.667 cm (infront of the lens)

now object distance for 2nd lens = 17+2.667 = 19.667 cm

1/f = 1/u + 1/v

1/-8 = 1/19.667 + 1/v

v = 5.6867 cm

b) The focal length, "fd" of the lens of diopter +2.1 for the farsighted man = -(1/2.1) = -0.476 m

The corrected focal length, F of the eye if uncorrected is f are related by

1/F = 1/f + 1/fd -0.03/(f*fd)----------------------------... 1

If fc is the focal length of contact lens, we have

1/F = 1/f + 1/fc ----------------------------------------... 2

Equating RHS of 1 and 2 we get

1/fc = 1/fd -0.03/(f*fd)----------------------------... 3

Substituting F= 0.28 and fd = -0.476 in 1 we get

1/0.28 = 1/f - 1/0.4762+ 0.03/(0.4762*f),

3.5714 = 1/f - 2.1 + 0.063/f

5.6714 = 1/f*(1+0.063)

1/f = 5.3353

f = 0.187

1/fc = 1/fd -0.03/(f*fd)

1/fc = 1/-0.4762+{0.03/(0.187*0.4762)} = -2.1 + 0.336 = -1.7638

fc = -0.567 m = 56.7 cm

diverging lens of +1.7638 diopter

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