[22]The defective eye of a person has a near point 0.5 m and far point 3 m. The power for corrective lenses required for (i) reading purpose and (ii) seeing distant objects respectively are?

The defective eye of a person has near point

0.5 m

and point

3 m

. The power for corrective lens required for reading purpose is

+ 2 D

and for seeing distant objects is

- 1 / 3 D

.
For reading purpose:

u = - 25 cm; v = 0.5 m = 50 cm; f = ?; P = ?

Using 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

)

\frac{1}{f} = \frac{1}{- 50} - \frac{1}{- 25}

\Rightarrow \frac{1}{f} = \frac{1}{50}

Therefore,

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

\Rightarrow P = \frac{100}{50}

\Rightarrow P = + 2 D

For distant objects:

u = \infty; v = - 3 m; f = ?; P = ?

\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

\frac{1}{f} = \frac{1}{- 3} - \frac{1}{\infty}

\Rightarrow \frac{1}{f} = - \frac{1}{3}

∴ P =

\frac{1}{f (m)}

\Rightarrow P = - \frac{1}{3} D

[22]The defective eye of a person has a near point $0.5 m$ and far point $3 m$ . The power for corrective lenses required for (i) reading purpose and (ii) seeing distant objects respectively are?