Question

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

Answer 1

the eye defect called Hypermetropia is corrected by using convex lens spectacles.

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



Object distance, u = -25 cm (Normal near point)

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

Focal length, f=? (To be calculated)

Putting these values in the lens formula,

By putting in values

• We get :  1/-100 - 1/(-25)= 1/f

Or,  -(1/100) + 1/25=1/f

-1+4/100= 1/f

3/100= 1/f

f= 100/3

f=33.3cm



Power of Lens needs to be calculated, Given by P=

∴ We convert 33.3 cm into meter i.e 0.33mP=

= +3.0 DThus, the power of the convex lens required is +3.0 diopters.

Answer 2

$\huge \bold \color{green} \mid\star { \underline { \underline{ \overline \red{answer}}}} \star\mid$

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...

$\huge \color{green} \star { \underbrace \red {\mathfrak{solved}} } \star$