Figure shows two spherical surface of radii R and 2R separating three transparent media of refractive index μ =4 , μ =2 and μ =1 . A ray of light travelling in a medium μ =1 is incident on outer sphere tangentially. The net deviation suffered by light ray will be

Question

Figure shows two spherical surface of radii R and 2R separating three transparent media of refractive index μ =4 , μ =2 and μ =1 . A ray of light travelling in a medium μ =1 is incident on outer sphere tangentially. The net deviation suffered by light ray will be <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-4iurcsn92j5o.png" /> (A) 0°  (B) 90°  (C) 240°  (D) 280°

Tardigrade Admin · Accepted Answer

Path of light ray for minimum deviation will be
Solution

s in 9 0^{\circ} = 2 s in θ

\Rightarrow θ = 3 0^{\circ}

Deviation

= 6 0^{\circ}

There are

4

such deviations therefore total deviation

= 24 0^{\circ}

.

Q. Figure shows two spherical surface of radii $R$ and $2 R$ separating three transparent media of refractive index $μ = 4$ , $μ = 2$ and $μ = 1$ . A ray of light travelling in a medium $μ = 1$ is incident on outer sphere tangentially. The net deviation suffered by light ray will be

$0^{\circ}$

$9 0^{\circ}$

$24 0^{\circ}$

$28 0^{\circ}$

Path of light ray for minimum deviation will be

$s in 9 0^{\circ} = 2 s in θ$
$\Rightarrow θ = 3 0^{\circ}$
Deviation $= 6 0^{\circ}$
There are $4$ such deviations therefore total deviation $= 24 0^{\circ}$ .

Q. Figure shows two spherical surface of radii R and 2R separating three transparent media of refractive index μ=4 , μ=2 and μ=1 . A ray of light travelling in a medium μ=1 is incident on outer sphere tangentially. The net deviation suffered by light ray will be

0∘

90∘

240∘

280∘