Question

A person uses a lens of power -2.0D for correcting his distant vision and for correcting his near vision he used lens of power +2.0D . Calculate the focal lenght of lenses required to correct these defects.

Answer 1

F=1/p
F=-1/2.0
F=-0.5 concave lens
F=1/p
F=1/2.0
F=0.5 convex lens

Answer 2

The focal length of lenses required to correct these defects are (-50 cm) and 50 cm respectively.

Explanation:

Given that,

Power used to correct distant vision, $P_1=-2\ D$

Power used to correct near vision, $P_2=+2\ D$

We need to find the focal length of lenses required to correct these defects. We know that the focal length is given by :

For first lens,

$f_1=\dfrac{1}{P_1}\\\\f_1=\dfrac{1}{-2\ D}\\\\f_1=-0.5\ m\\\\f_1=-50\ cm$

For second lens,

$f_2=\dfrac{1}{P_2}\\\\f_2=\dfrac{1}{2\ D}\\\\f_2=0.5\ m\\\\f_2=50\ cm$

So, the focal length of lenses required to correct these defects are (-50 cm) and 50 cm respectively. Hence, this is the required solution.

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