Optics Ques 51
- 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: 51.(c)
Solution:
Formula:
Critical Angle and Total Internal Reflection (T. I. R.)
- 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 $
BETA
