Question

A person cannot see objects beyond 80 cm from his eye while a person with normal eyesight can see object easily placed upto 160 cm from the eye. Find the nature, the focal length and the power of the correcting lens.

Answer 1

The concave lens of focal length 1.6 m and power 0.625 D should be used.

Explanation:

To calculate the focal length use lens formula as,

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

Here, v is image distance and u is the object distance.

Given $u=-160cm$ and $v=-80cm$.

Substitute the value, we get

$\frac{1}{f}=\frac{1}{-80} -\frac{1}{-160}$

$\frac{1}{f}=\frac{1}{-160}$

f = -160 cm.

Thus, focal length is -1.60 m.

Power of the lens,

$P=\frac{1}{f(m)} =\frac{1}{1.6} =0.625 D$.

Thus, the power of the lens is  0.625 D.

Hence, concave lens of focal length 1.6 m and power 0.625 D should be used.

#Learn More: lens formula

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Answer 2

Answer:

Explanation:

As a person cannot see far objects , i.e beyond 80 cm, he is suffering from myopia.

Hence, by using lens formula, we can find the focal length and power of a correcting lens.

Given :-

u = - 160 cm

v = - 80 cm

Solution :-

Using, 1/f = 1/v - 1/u, we get

⇒ 1/f = 1/- 80 + 1/160

⇒ 1/f = - 2 + 1/160

⇒ 1/f = - 1/160

⇒ f = - 160 cm = 1.6 m

Hence, the focal length is - 160 cm or 1.6 m.

Power, P = 1/f(in metre)

⇒ P = 1/1.6

⇒ P = - 0.625 D

Hence, the power of the correcting lens is  - 0.625 D.