Question

A ray of light is travelling from a medium of refractive index 1.414 to air. If the angle of incidence is $48^{\circ}$, then the angle of deviation of the light ray is

• A. $84^{\circ}$
• B. $42^{\circ}$
• C. $68^{\circ}$
• D. $102^{\circ}$

Hint:

Always calculate the critical angle first when light travels from denser to rarer medium. If $i<C$, refraction occurs ($\delta = r - i$). If $i>C$, reflection occurs ($\delta = 180 - 2i$).

Answer 1

The Correct Option is A

Step 1: Check for Total Internal Reflection (TIR):
Refractive index $\mu = 1.414 \approx \sqrt{2}$. Critical angle $C = \sin^{-1}(1/\mu) = \sin^{-1}(1/\sqrt{2}) = 45^{\circ}$. Given angle of incidence $i = 48^{\circ}$. Since $i>C$ ($48^{\circ}>45^{\circ}$), Total Internal Reflection occurs.
Step 2: Calculate Deviation:
When TIR occurs, the light reflects back into the same medium following the law of reflection ($i = r$). The angle of deviation $\delta$ is the angle between the extended incident ray and the reflected ray. \[ \delta = 180^{\circ} - 2i \]
Step 3: Calculation:
\[ \delta = 180^{\circ} - 2(48^{\circ}) \] \[ \delta = 180^{\circ} - 96^{\circ} \] \[ \delta = 84^{\circ} \]
Step 4: Final Answer:
The angle of deviation is $84^{\circ}$.