[5]The near point of a hypermetropic eye is 1 m. What is the power of the lens required to correct this defect? (Assume that the near point of the normal eye is 25 cm.)

The near point of a hypermetropic eye is

1 m

. The power of the lens required to correct this defect is

+ 3 D

.
To correct this defect, the image of an object

25 cm

should be brought to

100 cm

.
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 in

cm

\Rightarrow \frac{1}{f} = \frac{1}{- 100} - \frac{1}{- 25}

\Rightarrow \frac{1}{f} = \frac{1}{- 100} + \frac{1}{25}

\Rightarrow \frac{1}{f} = \frac{- 1 + 4}{100}

We get,

\Rightarrow f = + \frac{100}{3}

\Rightarrow f = + 33.3 cm

.
So, a convex lens of focal length

+ 33.3 cm

is required. (as

f

is positive, the lens used will be a convex lens)

Power = \frac{100}{f (cm)}

= + \frac{100}{33.3}

= + 3 D

[5]The near point of a hypermetropic eye is $1 m$ . What is the power of the lens required to correct this defect? (Assume that the near point of the normal eye is $25 cm$ .)