JEE Main 2016 · 3 Apr · Shift 1 · Physics Q28

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PhysicsSingle correct+4 / -1Question 28 of 90

In an experiment for determination of refractive index of glass of a prism by $i-\delta$ , plot, it was found that a ray incident at angle 35°, suffers a deviation of 40° and that it emerges at angle 79°. In that case which of the following is closest to the maximum possible value of the refractive index ?

1.5

1.6

1.7

1.8

Answer and worked solution

Official answer: A

$A=i+e-\delta=35+79-40=74^\circ$ . The minimum deviation cannot exceed 40°, so
$\mu\le\frac{\sin\frac{74^\circ+40^\circ}2}{\sin37^\circ}=\frac{\sin57^\circ}{\sin37^\circ}\approx1.39.$
Of the options, the closest is 1.5.

Topics:Prism and dispersion

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JEE Main 2016 · 3 Apr · Shift 1, Physics question 28