Question

A hypermetropic person with near point 1 m wants to
read newspaper keeping it at a distance of 25 cm. The
power of the corrective lens should be..​

Answer 1

Answer:

Hi hope u get help

Explanation:

u=-25cm

v=-100cm

f=??

1/f=1/v-1/u

1/f=-1/100+1/25

1/f=3/100

f=100/3 cm or 1/3 m

Now,

P=1/f

1/1/3=3D

Hence power is +3D

PLZ MARK BRAINLIEST

Answer 2

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In the problem, it is given that the near point of defective eye is 1 m and that of a normal eye is 25 cm. Hence u=−25 cm. The lens used forms its virtual image at near point of hypermetropic eye i.e., v=−1m=−100 cm. 

Using lens formula, we have :-

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

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

$f = \frac{100}{3} = 0.33m$

$power = \frac{1}{f(in \: metres)} = + 3.0d$