A glass slab consists of thin uniform layers of progressively decreasing Refractive Indices (RI) (see figure) such that the RI of mth layer is µ – m Δµ. Here, µ and Δµ denote the RI of 0th layer and the difference in RI between any two consecutive layers, repectively. The integer m = 0, 1, 2, 3, .... denotes the number of the successive layers. A ray of light from the 0th layer enters the 1st layer at an angle of incidence of sin–1 (7/10). After undergoing the mth refraction, the ray emerges parallel to the interface. If µ = 1.5 and Δµ = 0.015, the value of m is

Engineering

Physics

Reflection at Plane Surface

Question

A glass slab consists of thin uniform layers of progressively decreasing Refractive Indices (RI) (see figure) such that the RI of m^th layer is µ – m Δµ. Here, µ and Δµ denote the RI of 0th layer and the difference in RI between any two consecutive layers, repectively. The integer m = 0, 1, 2, 3, .... denotes the number of the successive layers. A ray of light from the 0th layer enters the 1st layer at an angle of incidence of sin^–1 (7/10). After undergoing the mth refraction, the ray emerges parallel to the interface. If µ = 1.5 and Δµ = 0.015, the value of m is

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By Snell's law, µ sin i = constant
µ sin i = (µ – mΔµ) sin r
where, µ = 1.5, i = 30°, r = 90°, Δµ = 0.015
$\frac{3}{2} \times \frac{1}{2} = (1.5 - m \times 0.015) \times 1$
$\frac{3}{4} = \frac{3}{2} \frac{15 m}{1000}$
$15 m = (\frac{3}{2} - \frac{3}{4}) \times 1000$
$15 m = \frac{6 - 3}{4} \times 1000 \Rightarrow m = \frac{3}{4} \times \frac{1000}{15}$
$r= \frac{3000}{60} \Rightarrow m = 50$

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