Q108 / 150 · MHT CET 2026 (19 April Shift 2) · Physics
Light travels a distance 'x' in time '$t_0$' in air and '$4x$' in time '$t_1$' in another denser medium. The critical angle for this medium is
- A$\sin^{-1}\left(\dfrac{t_1}{t_0}\right)$
- Correct answerB$\sin^{-1}\left(\dfrac{4t_0}{t_1}\right)$
- C$\sin^{-1}\left(\dfrac{4t_1}{t_0}\right)$
- D$\sin^{-1}\left(\dfrac{t_0}{4t_1}\right)$
Solution
Speed of light in air is given by $v_1 = \dfrac{x}{t_0}$
Speed of light in the denser medium is given by $v_2 = \dfrac{4x}{t_1}$
The refractive index of the denser medium with respect to air is $\mu = \dfrac{v_1}{v_2}$
The critical angle $\theta_c$ for the medium is given by $\sin \theta_c = \dfrac{1}{\mu} = \dfrac{v_2}{v_1}$
Substituting the values of $v_1$ and $v_2$:
$\sin \theta_c = \dfrac{\dfrac{4x}{t_1}}{\dfrac{x}{t_0}} = \dfrac{4t_0}{t_1}$
$\theta_c = \sin^{-1}\left(\dfrac{4t_0}{t_1}\right)$
Answer: $\sin^{-1}\left(\dfrac{4t_0}{t_1}\right)$
Speed of light in the denser medium is given by $v_2 = \dfrac{4x}{t_1}$
The refractive index of the denser medium with respect to air is $\mu = \dfrac{v_1}{v_2}$
The critical angle $\theta_c$ for the medium is given by $\sin \theta_c = \dfrac{1}{\mu} = \dfrac{v_2}{v_1}$
Substituting the values of $v_1$ and $v_2$:
$\sin \theta_c = \dfrac{\dfrac{4x}{t_1}}{\dfrac{x}{t_0}} = \dfrac{4t_0}{t_1}$
$\theta_c = \sin^{-1}\left(\dfrac{4t_0}{t_1}\right)$
Answer: $\sin^{-1}\left(\dfrac{4t_0}{t_1}\right)$
Hence, the correct answer is option: 'B'.