Question

Consider light travelling from a medium A to medium B separated by a plane interface. If the light undergoes total internal reflection during its travel from medium A to B and the speed of light in media A and B are \(2.4 \times 10^8\) m/s and \(2.7 \times 10^8\) m/s, respectively, then the value of critical angle is :

• A. sin⁻¹(9/8)
• B. cos⁻¹(8/9)
• C. tan⁻¹(8/√17)
• D. cot⁻¹(3/√15)

Hint:

If you know $\sin \theta = a/b$, you can always find other trigonometric ratios by drawing a right-angled triangle. This is a common trick in JEE options.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Total Internal Reflection (TIR) occurs when light travels from a denser medium to a rarer medium. The critical angle $\theta_c$ is the angle of incidence for which the angle of refraction is $90^\circ$.

Step 2: Key Formula or Approach:
1. Refractive index $\mu = \frac{c}{v}$.
2. $\sin \theta_c = \frac{\mu_{rarer}}{\mu_{denser}} = \frac{v_{denser}}{v_{rarer}}$.
Step 3: Detailed Explanation:
Given $v_A = 2.4 \times 10^8$ m/s and $v_B = 2.7 \times 10^8$ m/s.
Since $v_A<v_B$, medium A is denser and medium B is rarer.
\[ \sin \theta_c = \frac{v_A}{v_B} = \frac{2.4 \times 10^8}{2.7 \times 10^8} = \frac{24}{27} = \frac{8}{9} \] To find $\theta_c$ in terms of $\tan^{-1}$: If $\sin \theta_c = \frac{8}{9}$, then opposite $= 8$ and hypotenuse $= 9$.
Adjacent side $= \sqrt{9^2 - 8^2} = \sqrt{81 - 64} = \sqrt{17}$.
\[ \tan \theta_c = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{8}{\sqrt{17}} \] \[ \theta_c = \tan^{-1}\left(\frac{8}{\sqrt{17}}\right) \]
Step 4: Final Answer:
The value of the critical angle is tan⁻¹(8/√17).