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  4. The radius of curvature of each surface of a convex lens of refractive index 1.5 is 40 cm. Its power is

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Q. The radius of curvature of each surface of a convex lens of refractive index $1.5$ is $40 \,cm$. Its power is

Ray Optics and Optical Instruments

Solution:

Power, $P = \frac{1}{f} =(\mu-1)[\frac{1}{R_1}-\frac{1}{R_2}]$
Here, $\mu = 1.5, R_1 = 40\, cm = 0.4\, cm$,
$ R_2 = - 0.4 \,m$
$P =(1.5 -1)(\frac{1}{0.4}+ \frac{1}{0.4}) $
$= 2.5\,D$