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[4]
The far point of a myopic person is $80 cm$ in front of the eye. What is the power of the lens required to correct the problem?

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- 1.25 D

+ 1.25 D

- 1.5 D

+ 1.5 D

answer is A.

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Detailed Solution

The far point of a myopic person is

80 cm

in front of the eye. The power of the lens required to correct the problem is

- 1.25 D .

The far point of a typical eye is infinity. The defective eye's far point is given as

80 cm

The eye is short-sighted. To correct this defect, one should bring the object from infinity to

80 cm

.
From the data given,

u = -

v = - 80 cm

f = ?

Substituting in lens formula,

\frac{1}{f} = \frac{1}{v} - \frac{1}{u}

where

u =

the distance between the object and the pole of the mirror,

v =

the distance between the image and the pole of the mirror,

f =

focal length
We get,

\frac{1}{f} = \frac{1}{(- 80)} - \frac{1}{(- \infty)}

\Rightarrow \frac{1}{f} = \frac{1}{(- 80)} + 0

(as

1 / \infty = 0

)

\Rightarrow f = - 80 cm

We know that

Power of the lens = \frac{100}{f (in cm)}

\Rightarrow P = \frac{100}{- 80}

\Rightarrow P = - 1.25 D

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