Optics 4 Question 3
3. In an experiment for determination of refractive index of glass of a prism by $i-\delta$ plot, it was found that a ray incident at an angle $35^{\circ}$ suffers a deviation of $40^{\circ}$ and that it emerges at an angle $79^{\circ}$. In that case, which of the following is closest to the maximum possible value of the refractive index?
(2016 Main)
(a) 1.5
(b) 1.6
(c) 1.7
(d) 1.8
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Answer:
Correct Answer: 3. (a)
Solution:
$$ \delta=\left(i _1+i _2\right)-A $$
$\Rightarrow \quad 40^{\circ}=\left(35^{\circ}+79^{\circ}\right)-A$
$$ \Rightarrow \quad A=74^{\circ} $$
Now, we know that
$$ \mu=\frac{\sin \Big(\frac{A+\delta _m}{2}\Big)}{\sin \Big(\frac{A}{2}\Big)} $$
It we take the given deviation as the minimum deviation then,
$$ \mu=\frac{\sin \Big(\frac{74^{\circ}+40^{\circ}}{2}\Big)}{\sin \Big(\frac{74^{\circ}}{2}\Big)}=1.51 $$
The given deviation may or may not be the minimum deviation. Rather it will be less than this value. Therefore, $\mu$ will be less than 1.51.
Hence, maximum possible value of refractive index is 1.51 .
