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Question: The critical angle for glass is \(41^{\circ}48\prime\) and that for water is \(48^{\circ}36\prime\)....

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

Explanation

Solution

Critical angle and the refractive index of the material are related bysin⁡C=1μ=vc\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=1μ=vc\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 CCif the angle of the refracted ray lies on the boundary of the surface i.e. angle of refraction is 90∘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 μ\muis the ratio of the speed of light in medium vv to speed in a vacuumcc.
sin⁡C=1μ=vc\sin C=\dfrac{1}{\mu}=\dfrac{v}{c}
Given that critical angle for glass Cg=41∘48′=41.8∘C_{g}=41^{\circ}48\prime=41.8^{\circ} and that for water Cw=48∘36′=48.6∘C_{w}=48^{\circ}36\prime=48.6^{\circ}.
Then, the refractive index of glassμg=1sin⁡Cg=1sin⁡(41.8)=10.666=1.50\mu_{g}=\dfrac{1}{\sin C_{g}}=\dfrac{1}{\sin(41.8)}=\dfrac{1}{0.666}=1.50
Similarly, the refractive index of waterμw=1sin⁡Cw=1sin⁡(48.6)=10.750=1.33\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 μgw=μgμw=1.501.33=1.22\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⁡Cgw=1μgw\sin C_{gw}=\dfrac{1}{\mu_{gw}}
Cgw=sin⁡−1(1μgw)=sin⁡−1(11.12)=sin⁡−1(0.886)=62.45∘=62∘28′{{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∘28′62^{\circ}28\prime

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