Optics Ques 56
- 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
(2019 Main, 10 April I)
(a) $\frac{2 R}{\mu _1-\mu _2}$
(b) $\frac{R}{2-\left(\mu _1-\mu _2\right)}$
(c) $\frac{R}{2\left(\mu _1-\mu _2\right)}$
(d) $\frac{R}{\mu _1-\mu _2}$
Show Answer
Answer:
Correct Answer: 56.(d)
Solution:
Formula:
- Focal length of a lens is given as
$$ \frac{1}{f}=\left(\mu _1-1\right) \frac{1}{R _1}-\frac{1}{R _2} $$
$\therefore$ Focal length of plano-convex lens, i.e. lens 1,
$$ R _1=\infty \text { and } R=-R $$
$$ \begin{aligned} & \Rightarrow \quad \frac{1}{f _1}=\frac{\mu _1-1}{R} \\ & \text { or } \quad f _1=\frac{R}{\left(\mu _1-1\right)} \end{aligned} $$
Similarly, focal length of plano-concave lens, i.e. lens 2,
$$ R _1=-R \text { and } R _2=\infty $$
$$ \begin{aligned} & \Rightarrow \quad \frac{1}{f _2}=-\frac{\left(\mu _2-1\right)}{R} \\ & \text { or } \quad f _2=\frac{-R}{\left(\mu _2-1\right)} \end{aligned} $$
From Eqs. (i) and (ii), net focal length is
$$ \begin{aligned} \quad \frac{1}{f} & =\frac{\mu _1-1}{R}-\frac{\mu _2-1}{R}=\frac{\mu _1-\mu _2}{R} \\ \Rightarrow \quad f & =\frac{R}{\mu _1-\mu _2} \end{aligned} $$
BETA
