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

The person has near point 100cm. It is needed to read at a distance of 20cm.

(a) When contact lens is used, u = – 20 cm = – 0.2m, v = – 100 cm = –1 m So, 1/f = 1/v – 1/u = 1/-1 – 1/-0.2 = - 1 + 5 = + 4D (b) When spectacles are used, u = – (20 – 2) = – 18 cm = – 0.18m, v = – 100 cm = –1 m So, 1/f = 1/v – 1/u = 1/-1 – 1/-0.18 = - 1+ 5.55 = + 4.5D

hope this helps!

Answer 2

Given :

near point at =100 cm

power of lens is needed to read at 20 cm if he/she uses

(a) contact lens:

u = – 20 cm = – 0.2 m,

v = – 100 cm = – 1 m

The lens formula is given by

1/v-1/u=1/f

On substituting the values in the above formula, we get:

1/f=1/-1-1/-0.2

=4

∴ Power of the lens, P = 1/f= 4 D

(b) When the person wears spectacles at a distance of 2 cm from the eyes:

u = – (20 – 2) = – 18 cm = – 0.18 m,

v = – 100 cm = –1 m

The lens formula is given by

1/v-1/u=1/f

On substituting the values, we get:

1/f=1/-1-1/-0.18

=4.53

∴ Power of the lens, P = 1/f= 4.53 D