A ray of light passes through four transparent media with refractive indices $μ_{1}$ , $μ_{2}$ , $μ_{3}$ and $μ_{4}$ as shown in the figure. The surfaces of all media are parallel. If the emergent ray $C D$ is parallel to the incident ray $A B$ , we must have [2001]

Question

Smart Achievers · Accepted Answer

(4)

Applying Snell's law at A,

${}^{1}μ_{2} = \frac{\sin i}{\sin r_{1}} = \frac{μ_{2}}{μ_{1}}$ $...(i)$

Applying Snell's law at B,

${}^{2}μ_{3} = \frac{\sin r_{1}}{\sin r_{2}} = \frac{μ_{3}}{μ_{2}}$ $...(ii)$

Again applying Snell's law at C,

${}^{3}μ_{4} = \frac{\sin r_{2}}{\sin i} = \frac{μ_{4}}{μ_{3}}$ $...(iii)$

Multiplying eq. (i), (ii) and (iii),

$\frac{μ_{2}}{μ_{1}} \times \frac{μ_{3}}{μ_{2}} \times \frac{μ_{4}}{μ_{3}} = \frac{\sin i}{\sin r_{1}} \times \frac{\sin r_{1}}{\sin r_{2}} \times \frac{\sin r_{2}}{\sin i}$

$∴$ $\frac{μ_{4}}{μ_{1}} = 1$ $∴$ $μ_{1} = μ_{4}$

If the emergent ray is parallel to the incident ray after travelling through a number of parallel interfaces, then the refractive index of the first and the last medium is always the same.

Here, the incident ray is medium-1 $O A ∥ C D$ , so the emergent ray is in medium-4.

$∴$ $μ_{1} = μ_{4}$

Solution

Q.
A ray of light passes through four transparent media with refractive indices $μ_{1}$ , $μ_{2}$ , $μ_{3}$ and $μ_{4}$ as shown in the figure. The surfaces of all media are parallel. If the emergent ray $C D$ is parallel to the incident ray $A B$ , we must have [2001]

1 $μ_{1} = μ_{2}$

2 $μ_{2} = μ_{3}$

3 $μ_{3} = μ_{4}$

4 $μ_{4} = μ_{1}$

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Solution

Q. A ray of light passes through four transparent media with refractive indices μ1, μ2, μ3 and μ4 as shown in the figure. The surfaces of all media are parallel. If the emergent ray CD is parallel to the incident ray AB, we must have [2001]

1 μ1=μ2

2 μ2=μ3

3 μ3=μ4