Question

(a) The power of a concave lens used for correcting a myopic eye is – 0.6 D. Find the far point of the eye.(b) The power of a convex lens used to correct a hypermetropic eye is + 0.8 D. Find the near point of the eye.Assume the least distance of distinct vision to be 25 cm.

Answer 1

Answer:

(ii) The power of a convex lens used to correct a hypermetropic eye is + 0.8 D. Find the near point of the eye. Assume the least distance ...

Answer 2

(a). The far point of the eye is 21.7 cm.

(b). The near point of the eye is 31.3 cm.

Explanation:

Given that,

Power = -0.6 D

(a). We need to calculate the focal length

Using formula of power

$P=\dfrac{100}{f}$

$f=\dfrac{100}{P}$

Put the value into the formula

$f=\dfrac{100}{-0.6}$

$f=-166.6\ m$

We need to calculate the image distance

Using formula of lens

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

Put the value into the formula

$\dfrac{1}{-166.6}=\dfrac{1}{v}+\dfrac{1}{25}$

$\dfrac{1}{v}=\dfrac{1}{-166.6}-\dfrac{1}{25}$

$\dfrac{1}{v}=-\dfrac{958}{20825}$

$v=-21.7\ cm$

The far point of the eye is 21.7 cm.

(b). If the power is 0.8 D

We need to calculate the focal length

Using formula of power

$P=\dfrac{100}{f}$

$f=\dfrac{100}{P}$

Put the value into the formula

$f=\dfrac{100}{0.8}$

$f=125\ m$

We need to calculate the image distance

Using formula of lens

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

Put the value into the formula

$\dfrac{1}{125}=\dfrac{1}{v}+\dfrac{1}{25}$

$\dfrac{1}{v}=\dfrac{1}{125}-\dfrac{1}{25}$

$\dfrac{1}{v}=-\dfrac{4}{125}$

$v=-31.3\ cm$

The near point of the eye is 31.3 cm.

Hence, (a). The far point of the eye is 21.7 cm.

(b). The near point of the eye is 31.3 cm.

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Topic : optics

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