Question

One plano-convex and one plano-concave lens of same radius of curvature $'R'$ but of different materials are joined side by side as shown in the figure. If the refractive index of the material of $1$ is $\mu_1$ and that of $2$ is $\mu_2$, then the focal length of the combination is :

• A. $\frac{R}{2-\left(\mu_{1} -\mu_{2}\right)} $
• B. $\frac{ 2 R}{\mu_{1} -\mu_{2} } $
• C. $\frac{R}{2\left(\mu_{1} -\mu_{2}\right)} $
• D. $\frac{ R}{\mu_{1} -\mu_{2} } $

Answer 1

Answer: (D)

For 1st lens $\frac{1}{f_{1}} = \left(\frac{\mu_{1} -1}{ 1}\right) \left(\frac{1}{\infty} - \frac{1}{-R}\right) =\frac{\mu_{1} -1}{ R}$
for 2nd lens $ \frac{1}{f_{2}} = \left(\frac{\mu_{2} -1}{1}\right) \left(\frac{1}{-R} -0\right) =- \frac{\mu_{2} -1}{R} $
$ \frac{1}{f_{eq}} = \frac{1}{f_{1}} + \frac{1}{f_{2}} $
$ \frac{1}{f_{eq}} = \frac{R}{\mu_{1}-1} + \frac{R}{-\left(\mu_{2} -1\right)} \Rightarrow \frac{1}{f_{eq}} = \frac{R}{\mu_{1} -\mu_{2}} $
Hence $ f_{eq} = \frac{\mu_{1} -\mu_{2}}{R} $