Optics Ques 114
- 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) $1.8$
Show Answer
Answer:
Correct Answer: 114.(a)
Solution:
Formula:
- $\delta=\left(i _1+i _2\right)-A$
$\Rightarrow \quad 40^{\circ}=\left(35^{\circ}+79^{\circ}\right)-A$
$ \Rightarrow \quad A=74^{\circ} $
Now, we know that
$ \mu=\frac{\sin (\frac{A+\delta _m}{2})}{\sin (\frac{A}{2})} $
It we take the given deviation as the minimum deviation then,
$ \mu=\frac{\sin (\frac{74^{\circ}+40^{\circ}}{2})}{\sin (\frac{74^{\circ}}{2})}=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 $1.51$ .
BETA
