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Q. A biconvex lens has a focal length $2/3$ times the radius of curvature of either surface. The refractive index of the lens material is

Ray Optics and Optical Instruments

Solution:

Using lens makers formula,
$\frac{1}{f} =\left(\mu-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$.
Here, $f =\frac{2}{3}R$,
$R_{1} = +R, R_{2}=-R $
$\therefore \frac{1}{\left(23\right)R} = \left(\mu-1\right)\left(\frac{1}{R}+\frac{1}{R}\right)$
$ =\frac{ \left(\mu-1\right)\times2}{R} $
$ \Rightarrow \mu = 1.75$