Question

Two media of refractive indices $n_1$ and $n_2$ have a plane interface. In the first medium, speed of light is $2.4\times10^8$ m/s and in the second medium, it is $2.8\times10^8$ m/s. Find the critical angle when light travels from 1st medium to 2nd medium:

• A. $\sin^{-1}\!\left(\dfrac{5}{7}\right)$
• B. $\sin^{-1}\!\left(\dfrac{1}{3}\right)$
• C. $\sin^{-1}\!\left(\dfrac{6}{7}\right)$
• D. $\sin^{-1}\!\left(\dfrac{1}{4}\right)$

Hint:

Always compare refractive indices first. Critical angle is defined only when light goes from optically denser to rarer medium and satisfies $\sin i_c=\dfrac{n_2}{n_1}$.

Answer 1

The Correct Option is C

Concept: Refractive index is related to speed of light by: \[ n=\frac{c}{v} \] For light going from a denser to a rarer medium, the critical angle $i_c$ is given by: \[ \sin i_c = \frac{n_2}{n_1} \quad (n_1>n_2) \]
Step 1: Calculate refractive indices of the two media. For first medium: \[ n_1=\frac{3\times10^8}{2.4\times10^8}=\frac{5}{4} \] For second medium: \[ n_2=\frac{3\times10^8}{2.8\times10^8}=\frac{15}{14} \] 
Step 2: Check condition for critical angle. Since $n_1>n_2$, critical angle exists when light travels from medium 1 to medium 2. 
Step 3: Apply the formula: \[ \sin i_c=\frac{n_2}{n_1} =\frac{\frac{15}{14}}{\frac{5}{4}} =\frac{15}{14}\cdot\frac{4}{5} =\frac{6}{7} \] Step 4: Hence, the critical angle is: \[ \boxed{i_c=\sin^{-1}\!\left(\dfrac{6}{7}\right)} \]