Question

A person needs a lens of power + 3D for correcting his near vision
and -3D for correcting his distant vision. Calculate the focal lengths
of the lenses required to correct these defects.​

Answer 1

The focal lengths of the lenses required to correct these defects are 33.3 cm and -33.3 cm respectively.

Explanation:

Given that,

Power of the lens for correcting near vision, $P_1 = +3\ D$

Power of the lens for correcting distant vision, $P_2 = -3\ D$

We need to find the focal lengths of the lenses required to correct these defects. We know that the reciprocal of focal length is called the power of a lens.

Focal length of the lens used to correct near vision

$f_1=\dfrac{1}{P_1}\\\\f_1=\dfrac{1}{3}\\\\f_1=0.333\ m\\\\f_1=33.3\ cm$

Focal length of the lens used to correct distant vision

$f_2=\dfrac{1}{P_2}\\\\f_2=\dfrac{1}{-3}\\\\f_2=-0.333\ m\\\\f_2=-33.3\ cm$

So, the focal lengths of the lenses required to correct these defects are 33.3 cm and -33.3 cm respectively.

Learn more,

Power of a lens

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