Question

A hypermetropic person has to use a lens of power +5 d to normalise his vision. The near point of the hypermetropic eye is

Answer 1

Answer:2.5

Explanation:

Answer 2

The near point of the hypermetropic eye is$-1m$

Step by step calculation:

Given data:

power of the lens$=+5d$

To find:

We must find the near point of the hypermetropia eye .

Formula used:

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

$f=$focal length

v$=$image distance

u$=$object distance

Solution:

The focal length is given as : $f=\frac{100}{power}$

$f=\frac{100}{3}$

f=$33.3$cm

using the mirror formula: $\frac{1}{f} =\frac{1}{v} -\frac{1}{u}$

$\frac{1}{33.3} =\frac{1}{v} -\frac{1}{-25}$

Where normal vision is $25cm$

$\frac{1}{v} =\frac{1}{33.3} -\frac{1}{25}$

$v=-100cm$

v=-1m

Therefore, the near point is $-1m$

#SPJ3