JEE Main · 2016 · 03 Apr 2016 · Offline · Geometrical Optics

In an experiment for determination of refractive index of glass of a prism by

i - δ

, plot, it was found that a ray incident at angle

3 5^{\circ}

, suffers a deviation of

4 0^{\circ}

and that it emerges at angle

7 9^{\circ}

. In that case which of the following is closest to the maximum possible value of the refractive index ?

Answer: (D)

Solution

The prism angle follows from

δ = i + e - A

, so

A = 3 5^{\circ} + 7 9^{\circ} - 4 0^{\circ} = 7 4^{\circ}

. The refractive index obeys

μ = \frac{s i n \frac{A + δ _{m i n}}{2}}{s i n \frac{A}{2}}

, and the smallest deviation along the

i

δ

curve is below the

4 0^{\circ}

seen here. Taking

δ_{m i n}

a little under

4 0^{\circ}

keeps

μ

below

\frac{s i n 5 7 ^{\circ}}{s i n 3 7 ^{\circ}} \approx 1.44

, so of the values offered the largest possible index is closest to 1.5.

JEE Main · 2016 · 03 Apr 2016 · Offline · Single correct option