Optics Ques 93
- A plano-convex lens (focal length $f _2$, refractive index $\mu _2$, radius of curvature $R$ ) fits exactly into a plano-concave lens (focal length $f _1$, refractive index $\mu _1$, radius of curvature $R$ ). Their plane surfaces are parallel to each other. Then, the focal length of the combination will be
(2019 Main, 12 Jan II)
(a) $f _1-f _2$
(b) $\frac{R}{\mu _2-\mu _1}$
(c) $f_1 + f_2$
(d) $\frac{2 f _1 f _2}{f _1+f _2}$
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Answer:
Correct Answer: 93.(c)
Solution:
Formula:
- Given combination is as shown below.
As lenses are in contact, the equivalent focal length of the combination is
$$ \frac{1}{f _{eq}}=\frac{1}{f _1}+\frac{1}{f _2} $$
Using lens Maker’s formula,
$$ \begin{alignedat} \frac{1}{f} & =(\mu-1) \frac{1}{R _1}-\frac{1}{R _2} \\ \text { Here, } \quad \frac{1}{f _1} & =\left(\mu _1-1\right) \frac{1}{-R}-\frac{1}{\infty}=\frac{\left(1-\mu _1\right)}{R} \\ \text { and } \quad \frac{1}{f _2} & =\left(\mu _2-1\right) \frac{1}{\infty}-\frac{1}{-R}=\frac{\left(\mu _2-1\right)}{R} \\ \therefore \quad \frac{1}{f _{eq}} & =\frac{\mu _1-1}{R}+\frac{1-\mu _2}{R} \\ & =\frac{\mu _2-1+1-\mu _1}{R}=\frac{\mu _2-\mu _1}{R} \end{aligned} $$
So, $\quad f _{eq}=\frac{R}{\mu _2-\mu _1}$.
BETA
