Question

The focal length of the plano convex lens is $2R$ where $R$ is the radius of curvature of the surface. Then the refractive index of the material of the lens is _______

Answer 1

Figure shows a plano-convex lens of refractive index $\mu$.

From Lens Maker's formula:-

$\frac{1}{f}=(\frac{{\mu }_2}{{\mu }_1}-1)(\frac{1}{R_1}-\frac{1}{R_2})$

where, ${\mu }_2$ and ${\mu }_1$ are refractive index of lens and surrounding medium respectively.

And, $R_1$ and $R_2$ are radius of 1st and 2nd surfaces respectively.

Here,

${\mu }_1=1,{\mu }_2=\mu ,R_1=-\infty ,R_2=-R,f=2R$

Hence,

$\frac{1}{2R}=(\mu -1)(\frac{1}{-\infty }-\frac{1}{-R})$

$⟹\frac{1}{2R}=(\mu -1)\frac{1}{R}$

$⟹\mu =1+\frac{1}{2}=1.5$

Hence, $\mu =1.5$