Optics 2 Question 5
5. A point source $S$ is placed at the bottom of a transparent block of height $10 mm$ and refractive index 2.72 . It is immersed in a lower refractive index liquid as shown in the figure. It is found that the light emerging from the block to the liquid forms a circular bright spot of diameter $11.54 mm$ on the top of the block. The refractive index of the liquid is
(2014 Adv.)
(a) 1.21
(b) 1.30
(c) 1.36
(d) 1.42
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Answer:
Correct Answer: 5. (c)
Solution:
- At point $Q$ angle of incidence is critical angle $\theta _C$, where
$$ \sin \theta _C=\frac{\mu _l}{\mu _{\text {block }}} $$
In $\triangle P Q S, \sin \theta _C=\frac{r}{\sqrt{r^{2}+h^{2}}}$
$\therefore \quad \frac{\mu _l}{\mu _{\text {block }}}=\frac{r}{\sqrt{r^{2}+h^{2}}}$
$\Rightarrow \quad \mu _l=\frac{r}{\sqrt{r^{2}+h^{2}}} \times 2.72$
$$ =\frac{5.77}{11.54} \times 2.72=1.36 $$
