Optics 3 Question 8
8. A plano-convex lens of refractive index $\mu _1$ and focal length $f _1$ is kept in contact with another plano-concave lens of refractive index $\mu _2$ and focal length $f _2$. If the radius of curvature of their spherical faces is $R$ each and $f _1=2 f _2$, then $\mu _1$ and $\mu _2$ are related as
(2019 Main, 10 Jan I)
(a) $3 \mu _2-2 \mu _1=1$
(b) $2 \mu _2-\mu _1=1$
(c) $2 \mu _1-\mu _2=1$
(d) $\mu _1+\mu _2=3$
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Answer:
Correct Answer: 8. (c)
Solution:
- Given,
$$ f _1=2 f _2 $$
$$ \Rightarrow \quad \frac{1}{\left|f _1\right|}=\frac{1}{\left|2 f _2\right|} $$
Using lens Maker’s formula, we get
$$ \begin{aligned} \frac{1}{f _1} & =\left(\mu _1-1\right) \frac{1}{R _1}-\frac{1}{R _2} \\ & \frac{1}{f _2}=\left(\mu _2-1\right) \frac{1}{R _1^{\prime}}-\frac{1}{R _2^{\prime}} \\ \Rightarrow & \left|\left(\mu _1-1\right) \frac{1}{\infty}-\frac{1}{-R}\right|=\left|\frac{1}{2}\left(\mu _2-1\right) \frac{1}{-R}-\frac{1}{\infty}\right| \end{aligned} $$
[ usingEq. (i)]
$$ \begin{array}{ll} \Rightarrow & \frac{\mu _1-1}{R}=\frac{\mu _2-1}{2 R} \\ \Rightarrow & 2 \mu _1-\mu _2=1 \end{array} $$
