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The correct answer is option 3) i.e. 80°. CONCEPT:
\(\sin c = \frac{\ 1}{\ \mu}\)
Refractive index (\(\mu\))= \(\frac{\sin i}{\sin r}\) Where i is the angle of incidence of light and r is the angle of refraction of light.
CALCULATION: Given that: Refractive index (\(\mu\))= 1.5 Angle of incidence (i) = 50º \(c = \sin ^{-1}\left(\frac{\ 1}{\ 1.5} \right )\)\(= 41.8^\circ\) \(\therefore\) \(i > c\) The given light ray, therefore, undergoes total internal reflection as shown: \(\angle ANO = \angle BNO = 50^\circ\) ( \(\because\) angle of incidence = angle of reflection) \(\angle ANO = \angle MNP = 50^\circ\) ( \(\because\) vertically opposite angle) Therefore deviation = difference between incident and reflected angles Deviation = 180° - (50° + 50°) = 80° Mistake Points Refractive index (\(\mu\))= 1.5 Angle of incidence (i) = 50º \(\mu = \frac{\sin 50^{\circ}}{\sin r} \Rightarrow \sin r = \frac{\sin 50^{\circ}}{\mu}\) \(\sin r = \frac{\sin 50^{\circ}}{1.5}\) \(r = \sin ^{-1}\left ( \frac{\sin 50^{\circ}}{1.5} \right )\)\(= 30.71^\circ\) Here the obtained result indicates r < i even though light enters from optically denser to optically rarer medium. This indicates that light has undergone a total internal reflection. India’s #1 Learning Platform Start Complete Exam Preparation
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