Question

The near point of a hypermetropic eye is 1.8 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.​

Answer 1

Given:-

• Image distance ,u = - 1.8m

• Object distance ,u = -25 cm

To Find:-

• Power of Lens ,P

Solution:-

Firstly we convert the unit here

1.8 m = 180 cm

The lens used to correct the defect is convex lens .

So by using lens Formula

• 1/v - 1/u = 1/f

Substitute the value we get

→ -1/180 - 1/(-25) = 1/f

→ -1/180 +1/25 = 1/f

→ -5+36/900 = 1/f

→ 31/900= 1/f

→ f = 900/31

→ f = 29.03 cm

Now , Calculating Power of Lens

As we know that Power of Lens is defined as the reciprocal of its focal length.

• P = 1/f (in metre)

Substitute the value we get

→ P = 100/29.02

→ P = 100/29.02

→ P = +3.44 D

Therefore, the power of lens is 3.44 Dioptre.

Answer 2

Given :-

☄ Image distance, v = - 1.8m = 180cm

☄ Object distance, u = -25 cm

To Find :-

Power of Lens to correct this defect

Solution:-

The lens used to correct the defect is convex lens.

using lens Formula .i.e.,

$\leadsto \tt \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{ 1}{f} \\</p><p>$

$\leadsto \tt \dfrac{ -1}{180 } - \dfrac{ 1}{(-25) }= \dfrac{1}{f}$

$\leadsto \tt \dfrac{ -1}{180} + \dfrac{1}{25} = \dfrac{1}{f}$

$\leadsto \tt</p><p> \dfrac{ -5+36}{900} = \dfrac{1}{f}$

$\leadsto \tt \dfrac</p><p>{31}{900} = \dfrac{1}{f}$

$\leadsto \tt </p><p> f = \dfrac{900}{31}$

$\leadsto \tt f = 29.02cm$

we know,

$\leadsto \tt P = \dfrac{1}{f}$

$\leadsto \tt</p><p> P = \dfrac{100}{29.02}$

$\leadsto \tt</p><p> P = +3.44 D$

thus, the power of lens is 3.44D (Dioptre).

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