# Optical instrument contact lenses – problems and solutions

1. Power of a contact lens is -5 Diopters.

(a) The contact lens is a converging lens or diverging lens?

(b) What is the focal length of the contact lens?

(c) Nearsighted eye or farsighted eye? If the nearsighted eye, what then will be the near point?

__Known :__

Lens’ power (P) = -5 D

__Solution :__

**(a) ****Converging lens or diverging lens ****? **

The minus sign of the lens’ power indicates the contact lens is the diverging lens.

**(b) **The focal length?** **

P = 1/f

-5 = 1/f

f = 1/-5 = -0.2 m = -20 cm

The focal length of diverging lens is 20 cm.

**(c) ****Nearsighted eye or farsighted eye ?**** **

-1/d_{i }= 1/f – 1/d_{o}

-1/d_{i} = -1/20 – 1/~ = -1/20 – 0

-1/d_{i} = -1/20

d_{i }= 20 cm

The image distance is 20 cm. 20 cm is far point of the farsighted eye. For a normal eye, the far point is infinity.

2. Power of a contact lens is

(a) The contact lens is converging lens or diverging lens ?

(b) What is the focal length of the contact lens ?

(c) Nearsighted eye or farsighted eye ? If farsighted eye, what then will be the far point ?

__Known :__

Lens power (P) = 1.5 Diopters

__Solution :__

**(a) ****converging lens or diverging lens**

The plus sign of the lens’ power indicates the contact lens is converging lens.

**(b) ****The focal length **** **

P = 1/f

1.5 = 1/f

f = 1/ 1.5 = 0.67 m = 67 cm

The focal length of converging lens is 67 cm.

**(c) ****Nearsighted eye or farsighted eye**

-1/d_{i} = 1/f – 1/d_{o}

-1/d_{i} = 1/67 – 1/25 = 25/1675 – 67/1675 = -42/1675

–d_{i }= -1675/42 = -40 cm = -0.40 m

d_{i }= 40 cm

The image distance is 40 cm. 20 cm is near point of the farsighted eye. For a normal eye, the near point is 25 cm.

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