Q105 / 150 · MHT CET 2021 (22 Sep Shift 2) · Physics
The critical angle for light going from medium ' $\mathrm{x}$ ' to medium ' $Y$ ' is $\theta$. The speed of light in medium ' $\mathrm{x}$ ' is ' $\mathrm{V}_{\mathrm{x}}$ '. The speed of light in medium ' $\mathrm{Y}$ ' is
- A$\mathrm{V}_{\mathrm{x}} \sin \theta$
- B$\mathrm{V}_{\mathrm{x}} \tan \theta$
- C$\frac{\mathrm{V}_{\mathrm{x}}}{\tan \theta}$
- Correct answerD$\frac{\mathrm{V}_{\mathrm{x}}}{\sin \theta}$
Solution
Refractive index of ' $x$ ' w.r.t. ' $y$ ' is
$$
\mathrm{y}_{\mathrm{x}}=\frac{1}{\sin \theta}
$$
Also ${ }_y n_x=\frac{V_y}{V_x}=\frac{1}{\sin \theta}$
$$
\therefore \mathrm{V}_{\mathrm{y}}=\frac{\mathrm{V}_{\mathrm{x}}}{\sin \theta}
$$
Hence, the correct answer is option: 'D'.