Question

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Try this .....NCERT CLASS 10 (Pg - 198)

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Answer 1

Hey user !!

Here's the answer you are looking for.

Since, the near point of the eye is 1m, the lens must make the objects kept at points closer than 1m to appear as if they are coming from 1m. So the virtual image of the object needs to be formed at 1m.

Basically, the lens formula,

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

u = -25cm

v = -100cm

So,

$\frac{1}{f} = \frac{ - 1}{100} - \frac{( - 1)}{25} \\ \\ \: \: \: \: = \frac{ - 1 + 4}{100} \\ \\ \: \: \: \: = \frac{3}{100} {cm }^{ - 1} \: = 3 {m}^{ - 1}$


$Power = \frac{1}{f} {m}^{ - 1} = \frac{1}{f} D \\ \\ Power = \: 3D$

★★ HOPE THAT HELPS ☺️ ★★

Answer 2

For Hypermetropic eye,

v=-1m=-100cm

u=-25cm
Using lens formula,
1/f=1/v -1/u
=1/(-100) - 1/(-25)
=3/100
or f=100 /3 cm=1/3 m
Now,
Power,P=1/f=3 D