Let the $x-z$ plane be the boundary between two transparent media. Medium 1 in $z \geq 0$ has a refractive index of $\sqrt{2}$ and medium 2 with $z<0$ has a refractive index of $\sqrt{3}$. A ray of light in medium 1 given by the vector $\vec{A}=6\sqrt{3}\hat{i}+8\sqrt{3}\hat{j}-10\hat{k}$ is incident on the plane of separation. The angle of refraction in medium 2 is

JEE Main · 2011 · 10 Apr 2011 · Offline · Geometrical Optics

Let the

x - z

plane be the boundary between two transparent media. Medium 1 in

z \geq 0

has a refractive index of

2

and medium 2 with

z < 0

has a refractive index of

3

. A ray of light in medium 1 given by the vector

A = 63 \hat{i} + 83 \hat{j} - 10 \hat{k}

is incident on the plane of separation. The angle of refraction in medium 2 is

Answer: (A)

Take the normal to the boundary along the

z

axis. The magnitude of the incident ray vector is

∣ A ∣ = (63)^{2} + (83)^{2} + 1 0^{2} = 108 + 192 + 100 = 20

, and its

z

component is

10

, so the angle of incidence satisfies

cos θ_{1} = \frac{10}{20} = \frac{1}{2}

, giving

θ_{1} = 6 0^{\circ}

. Snell's law across the interface,

μ_{1} sin θ_{1} = μ_{2} sin θ_{2}

, becomes

2 sin 6 0^{\circ} = 3 sin θ_{2}

, that is

2 \cdot \frac{3}{2} = 3 sin θ_{2}

, so

sin θ_{2} = \frac{1}{2}

and

θ_{2} = 4 5^{\circ}

JEE Main · 2011 · 10 Apr 2011 · Offline · Single correct option