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Q. Assertion : The focal length of an equiconvex lens of radius of curvature $R$ made of material of refractive index $\mu = 1.5$, is $R$.
Reason : The focal length of the lens will be $R/2$.

Ray Optics and Optical Instruments

Solution:

$\frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)=(1.5-1)\left(\frac{1}{R}-\frac{1}{-R}\right)$
or $f=R$.