Optics 3 Question 1

1. 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) $\frac{R}{\mu _1-\mu _2}$

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Answer:

Correct Answer: 1. (d)

Solution:

  1. 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

$$ \begin{aligned} \quad \frac{1}{f} & =\frac{\mu _1-1}{R}-\frac{\mu _2-1}{R}=\frac{\mu _1-\mu _2}{R} \\ \Rightarrow \quad f & =\frac{R}{\mu _1-\mu _2} \end{aligned} $$



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