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Q. Assertion : The focal length of an equiconvex lens placed in air is equal to radius of curvature of either face.
Reason: For an equiconvex lens radius of curvature of both the faces is same.

Ray Optics and Optical Instruments

Solution:

For an equiconvex lens, $R_1 = R_2 = R$
From $\frac{1}{f} = (\mu -1) (\frac{1}{R_1}-\frac{1}{R_2})$
[using lens makers formula]
For lens, $\mu = 1.5$ placed in air.
Therefore, $\frac{1}{f} = (1 .5 -1 )\frac{2}{R}$
$ \Rightarrow f = R$,