Question

The near point of a hypermetropic eye is 1m.What is the power of lens required to correct this defect?Assume the near point of the normal eye is 25cm.

Answer 1

hence the person suffering from hypermetropia should use a lens of power +0.03 that is a convex lens of focal length 33.3cm

Answer 2

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the eye defect called hypertrophic is corroded by using a convex lens..so, the person requires convex lens spectacle..

we will first calculate the focal length of the convex lens required in this case..this hypermetropic eye can see the nearby object kept at 25 cm (at near point of normal eye) clearly if the image of this object is formed at its own near point which is 1 metre here..so, in this case:-

object distance,u = -25cm (normal near point)

image distance, v = -1m = -100cm ( near point of this defective eye)

Focal length , f = ?? ( to be calculated)

now.. putting these values in the lens formula,

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

we get :

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

or

$\huge \bold{- \frac{1}{100} + \frac{1}{25} = \frac{1}{f} }$

$\huge \bold{ \implies \frac{ - 1 + 4}{100} = \frac{1}{f} }$

$\huge \bold{\implies \: \frac{1}{f} = \frac{3}{100} }$

$\huge \bold {\implies \: f = \frac{100}{3} }$

$\huge \bold{\: f = 33.3cm}$

thus, the focal length of the concave lens required is +33.3cm ..we will now calculate the follower, please note that 33.3cm is equal to 33.3/100m or 0.33m Now,

$\bold{power \: p \: = \frac{1}{f(in \: metres)} }$

$\implies \: p = \frac{1}{ + 0.33} \\ = + \frac{ 100}{3.3} \\ \\ = + 3.0D$

so, the power of convex lens required is +3.0 dioptres...

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