Question

A person wants to read a book placed at 15 cm, whereas near point of his eye is 30 cm. What is the required power of the lens?

Answer 1

Answer:

Explanation:confidence is the key(power in this case).

Hope

your doubt is clarified

Answer 2

Answer:

Explanation:

Let x be the distance near point and d be the distance of the object.

Now :

f = xd/(x - d)

In this case :

x = 30 cm

d = 20 cm

Doing the substitution we have :

f = (20 × 30)/(30 - 20)

f = 600/10

We now have f as follows.

f = 60

The power formula is given as follows :

P = 100/f

Doing the substitution we have :

P = 100/60

= 1.67D

This person is hypermetropic which means he is long-sighted .