Optics 3 Question 8

(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:

  1. 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)]

$$ \begin{array}{ll} \Rightarrow & \frac{\mu _1-1}{R}=\frac{\mu _2-1}{2 R} \\ \Rightarrow & 2 \mu _1-\mu _2=1 \end{array} $$



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