Question

The critical angle for glass is $41^{\circ}48\prime$ and that for water is $48^{\circ}36\prime$. Calculate the critical angle for glass-water interface.
A.$62^{\circ}28\prime$
B.$34^{\circ}42\prime$
C.$52^{\circ}42\prime$
D.$44^{\circ}42\prime$

Answer 1

Hint: Critical angle and the refractive index of the material are related by$\sin C=\dfrac{1}{\mu}=\dfrac{v}{c}$. Since the critical angle of two different systems is given, we can find the refractive index of each system and then the refractive index of the combined materials. This refractive index will give the required critical angle.

Formula used:
$\sin C=\dfrac{1}{\mu}=\dfrac{v}{c}$

Complete step-by-step answer:
We know that the angle of incidence is called the critical angle $C$if the angle of the refracted ray lies on the boundary of the surface i.e. angle of refraction is $90^{\circ}$. Also the sine of the critical angle gives the inverse of the refractive index of the material. We also know that the refractive index $\mu$is the ratio of the speed of light in medium $v$ to speed in a vacuum$c$.
$\sin C=\dfrac{1}{\mu}=\dfrac{v}{c}$
Given that critical angle for glass $C_{g}=41^{\circ}48\prime=41.8^{\circ}$ and that for water $C_{w}=48^{\circ}36\prime=48.6^{\circ}$.
Then, the refractive index of glass$\mu_{g}=\dfrac{1}{\sin C_{g}}=\dfrac{1}{\sin(41.8)}=\dfrac{1}{0.666}=1.50$
Similarly, the refractive index of water$\mu_{w}=\dfrac{1}{\sin C_{w}}=\dfrac{1}{\sin(48.6)}=\dfrac{1}{0.750}=1.33$
Then, from the definition of the refractive index, the refractive index of glass-water interface is $\mu_{gw}=\dfrac{\mu_{g}}{\mu_{w}}=\dfrac{1.50}{1.33}=1.22$
The critical of the of glass-water interface is given as $\sin C_{gw}=\dfrac{1}{\mu_{gw}}$
\[{{C}_{gw}}={{\sin }^{-1}}\left( \frac{1}{{{\mu }_{gw}}} \right)={{\sin }^{-1}}\left( \frac{1}{1.12} \right)={{\sin }^{-1}}(0.886)={{62.45}^{\circ }}={{62}^{\circ }}{{28}^{'}}\]
Hence the answer is A.$62^{\circ}28\prime$

Note:
 To calculate the critical angle we must take the $sin^{-1}$ of the reflective index and not $\dfrac{1}{\sin \theta}= \mathrm{cosec} \theta$, which are both different. Also note that the refractive index of glass-water interface is $\mu_{gw}=\dfrac{\mu_{g}}{\mu_{w}}$. The angle of incidence is called the critical angle $C$if the angle of the refracted ray lies on the boundary of the surface i.e. angle of refraction is $90^{\circ}$.