Question

A man can see only between 75 cm and 200 cm.The power of lens to correct the near point will be

Answer 1

Answer: +8/3 D

Explanation:

Here we are given that a man can see between 75 cm and 200 cm i.e., his near point = 75 cm and far point = 200 cm  

The least distance of distinct vision for a normal human eye = 25 cm

So, in order to correct the near point of the man we have to use a convex lens so that the object is at 25 cm and the image is obtained at 75 cm i.e.,  

Object distance, u = - 25 cm  

Image distance, v = - 75 cm

Now, by using the lens formula,  

$\frac{1}{f}$ = $\frac{1}{v}$ – $\frac{1}{u}$

⇒ $\frac{1}{f}$ = $\frac{1}{-75}$ – $\frac{1}{-25}$

⇒ $\frac{1}{f}$ = $\frac{1}{25}$  – $\frac{1}{75}$

⇒ $\frac{1}{f}$ =  $\frac{3-1}{75}$ = $\frac{2}{75}$

⇒ $\frac{1}{f}$ = $\frac{75}{2}$  cm = $\frac{75/2}{100}$ m ….. (i)

Thus,  

The power of the convex lens to correct the near point is,

= $\frac{1}{f}$ meter

= 100/[75/2] ….. [from (i)]

= 200 / 75  

= +8/3 D