MHT CET · Physics · Wave Optics
A plane wavefront is incident on a water surface at an angle of incidence \(60^{\circ}\) then it gets refracted at \(45^{\circ} .\) The ratio of width of incident wavefront to that of refracted
wavefront will be \(\left[\sin \frac{\pi}{4}=\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}, \sin 60^{\circ}=\frac{\sqrt{3}}{2}, \cos 60^{\circ}=\frac{1}{2}\right]\)
- A \(\frac{\sqrt{3}}{2}\)
- B \(2 \sqrt{3}\)
- C \(\frac{1}{\sqrt{2}}\)
- D \(\sqrt{2}\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{\sqrt{2}}\)
Step-by-step Solution
Detailed explanation
\( \frac{W_i}{W_r} = \frac{\cos \theta_i}{\cos \theta_r} \) \( = \frac{\cos 60^{\circ}}{\cos 45^{\circ}} \)
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