Question

A person has near point at 100 cm. What power of lens is needed to read at 20 cm if he/she uses (a) contact lens, (b) spectacles having glasses 2.0 cm separated from the eyes?

Answer 1

Explanation:

l think you should take help of your teacher they tell you more clearly

Answer 2

Lens of powers 4 D & 4.5 D can be Used

Explanation:

• The Near point for Person = 100 cm, So it should be read as 20 cm in terms of distance.

• a.) With contact lens in use-  

u = - 20 cm = - 0.2 m

v = - 100 cm = - 1 m

So, $\frac {1}{f} = \frac {1}{v} - \frac {1}{u}$

= $\frac {1}{-1} - \frac {1}{- 0.2}$

= - 1 + 5  

= + 4 D

• b.) When spectacles came into use -

u = - (20 - 2) = - 18 cm = - 0.18 m

v = - 100 cm = - 1 m

So,  $\frac {1}{f} = \frac {1}{v} - \frac {1}{u}$

= $\frac {1}{-1} - \frac {1}{- 0.18}$

= - 1 + 5.55  

= + 4.5 D