[3]
The far point of a myopic person is $80 cm$ in front of the eye. What is the nature of the lens required to correct the problem?

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Concave lens

Convex lens

Both

None

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 nature of the lens required to correct this problem is a concave lens.
The far point of a normal human eye is infinity. As the far point of the defective eye 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 = - \infty

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

Since the focal length required to treat the person’s defect is of negative sign, this shows that the lens needed for it is a Concave lens, as concave lens have a negative focal length.

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