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Q.
An optician prescribes spectacles to a patient with a combination of a convex lens of focal length $40\, cm$ and concave lens $25\, cm$. The power of spectacles is:
AFMCAFMC 2002
Solution:
Power of a lens is equal to inverse of its focal length.
The power of a thin lens is equal to the reciprocal of its focal length $(f)$ measured in meters
$P=\frac{1}{f \text { (meters) }}$
Given, focal length of convex lens $f_{1}=40\, cm$
and of concave lens $f_{2}=-25\, cm$
$\therefore $ Focal length of combination
$\frac{1}{F} =\frac{1}{f_{1}}+\frac{1}{f_{2}}$
$=\frac{1}{40}-\frac{1}{25}=-\frac{3}{200}$
$\Rightarrow F=-\frac{200}{3}=-66.7\, cm$
Power of spectacles is
$=\frac{100}{F}=\frac{100}{-66.7}=-1.5\, D$