Question

A perdon cannot see object nearer than 75cm from his eyes while a person with normal vision

Answer 1

The person is suffering from hypermetropia.

So, u=normal near point=-25cm

v=defective near point=-75cm

Using lens formula-

1/v-1/u=1/f

1/-75-(1/-25)=1/f

-1/75+1/25=1/f

-1+3/75=1/f

2/75=1/f

f=75/2

f=37.5cm

P=1/f

P=100/37.5 (as f is in cm)

P=1000/375

P= +2.66 D

Answer 2

$\huge{\underline{\underline{\red{\mathfrak{Correct \: Question :}}}}}$

A person cannot see object nearer than 75cm from his eyes while a person with normal vision can see objects Upto 25 cm from his eyes.Find the nature, the focal length and the power of correcting lens used for the defective vision.

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$\huge{\underline{\underline{\blue{\mathfrak{Answer :}}}}}$

The defect of the vision is Hypermetropia.

Given that,

• Image distance (v) = -75 cm

• Object Distance (u) = -25 cm

___________________________

We know that,

$\Large{\boxed{\boxed{\purple{\sf{\frac{1}{f} = \frac{1}{v} - \frac{1}{u}}}}}}$

(Putting Values)

$\sf{ \frac{1}{f} = \frac{ - 1}{75} - ( - \frac{1}{25}) } \\ \\ \sf{ \frac{1}{f} = \frac{ - 1}{75} + \frac{1}{25} } \\ \\ \sf{ \frac{1}{f} = \frac{ - 1 + 3}{75} } \\ \\ \sf{ \frac{1}{f} = \frac{2}{75} } \\ \\ \sf{f = \frac{75}{2} } \\ \\ \sf{f = 37.5 \: cm}$

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Now,

Power of lens = + 100/37.5

Power of lens = + 2.67 D

$\large{\boxed{\green{\bf{Power = 2.57 \: D}}}}$

The positive sign with f or the power indicates a convergent lens

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