Question

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

Answer 1

Answer:

Here, near point is 1 m.

Explanation:

Hypermetropia can be corrected by using a convex lens. A convex lens converges the incoming light such that the image is formed on the retina. An object at 25 cm forms an image at the near point of the hypermetropic eye.

Answer 2

Given :-

• Object distance (u) = -25 cm

• Image distance (v) = -1 m = -100 cm

To find :-

• Nature and power of lens required to correct hypermetropia

Concept :-

• The eye defect called hypermetropia and it is corrected by using convex lens.so,the person requires convex lens spectacles

• We will first calculate the focal length of convex lens required in this case.

Solution :-

$\green{\bigstar}\:\underbrace{\bf\orange{{By\:using\:lens\:formula}}}$

$\red{\bigstar}\:{\underline{\boxed{\bf\purple{\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}}}}}$

$\rm\longrightarrow\:\dfrac{1}{-100}-\dfrac{1}{-25}=\dfrac{1}{f}$

$\rm\longrightarrow\:-\dfrac{1}{100}+\dfrac{1}{25}=\dfrac{1}{f}$

$\rm\longrightarrow\:\dfrac{-1+4}{100}=\dfrac{1}{f}$

$\rm\longrightarrow\:\dfrac{3}{100}=\dfrac{1}{f}$

$\rm\longrightarrow\:f=\dfrac{100}{3}$

$\rm\longrightarrow\:f=33.3\:cm$

Hence the focal length of the convex lens required will be 33.3 cm.

Now,

$\blue{\bigstar}\:{\underline{\boxed{\bf\pink{p=\dfrac{1}{f\small(in\:metres)}}}}}$

$\rm\longrightarrow\:p=\dfrac{1}{0.33}\times\:100$

$\rm\longrightarrow\:p=\dfrac{100}{33}$

$\rm\longrightarrow\:p=3.03 \:D$

Hence,the power of convex lens required is 3.03 dioptres.