Optics Ques 55
- A ray of light is incident at the glass-water interface at an angle $i$, it emerges finally parallel to the surface of water, then the value of $\mu _g$ would be
$(2003,2 M)$
(a) $(4 / 3) \sin i$
(b) $1 /(\sin i)$
(c) $4 / 3$
(d) $1$
Show Answer
Answer:
Correct Answer: 55.(b)
Solution:
Formula:
Critical Angle and Total Internal Reflection (T. I. R.)
- Applying Snell’s law $(\mu \sin i=$ constant $)$
at $1$ and $2$ , we have
$\mu _1 \sin i _1=\mu _2 \sin i _2$
Here,
$ \begin{aligned} & \mu _1=\mu _{\text {glass }}, i _1=i \\ & \mu _2=\mu _{\text {air }}=1 \text { and } i _2=90^{\circ} \end{aligned} $
$\therefore \quad \mu _g \sin i=(1)\left(\sin 90^{\circ}\right)$ or $\mu _g=\frac{1}{\sin i}$
BETA
