MHT CET · Physics · Ray Optics
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 \(\frac{\mathrm{V}_{\mathrm{x}}}{\tan \theta}\)
- D \(\frac{\mathrm{V}_{\mathrm{x}}}{\sin \theta}\)
Answer & Solution
Correct Answer
(D) \(\frac{\mathrm{V}_{\mathrm{x}}}{\sin \theta}\)
Step-by-step Solution
Detailed explanation
Refractive index of ' \(x\) ' w.r.t. ' \(y\) ' is
\(
\mathrm{y}_{\mathrm{x}}=\frac{1}{\sin \theta}
\)
Also \({ }_y n_x=\frac{V_y}{V_x}=\frac{1}{\sin \theta}\)
\(
\therefore \mathrm{V}_{\mathrm{y}}=\frac{\mathrm{V}_{\mathrm{x}}}{\sin \theta}
\)
\(
\mathrm{y}_{\mathrm{x}}=\frac{1}{\sin \theta}
\)
Also \({ }_y n_x=\frac{V_y}{V_x}=\frac{1}{\sin \theta}\)
\(
\therefore \mathrm{V}_{\mathrm{y}}=\frac{\mathrm{V}_{\mathrm{x}}}{\sin \theta}
\)
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