Question

If ${}_i{\mu ^j}$ represents a refractive index when a light ray goes from medium $i$ to medium $j$. Then the product ${}_2{\mu ^1} \times {}_3{\mu ^2} \times {}_4{\mu ^3}$ is equals to
(A) ${}_3{\mu ^1}$
(B) ${}_3{\mu ^2}$
(C) ${}_4{\mu ^1}$
(D) ${}_4{\mu ^2}$

Answer 1

Hint: The product of the three refractive indexes of light from one medium to another medium is determined by using the refractive index formula from one medium to another medium. This formula shows the relation between the two refractive indices.

Formula used:
refractive index from one medium to another medium is given by,
${}_1{\mu ^2} = \dfrac{{{\mu _1}}}{{{\mu _2}}}$
Where, ${}_1{\mu ^2}$ is the refractive index of the light from medium $1$ to medium $2$, ${\mu _1}$ is the refractive index of the first medium, and ${\mu _2}$ is the refractive index of the second medium

Complete step by step answer:
Given that,
If ${}_i{\mu ^j}$ represents refractive index when a light ray goes from medium $i$ to medium $j$, then by using the refractive index from one medium to another medium formula,
${}_i{\mu ^j} = \dfrac{{{\mu _i}}}{{{\mu _j}}}$
Similarly, for ${}_2{\mu ^1}$, the refractive index from medium $2$ to medium $1$ , then by using the refractive index from one medium to another medium formula,
$\Rightarrow {}_2{\mu ^1} = \dfrac{{{\mu _2}}}{{{\mu _1}}}\,.......................\left( 1 \right)$
Similarly, for ${}_3{\mu ^2}$, the refractive index from medium $3$ to medium $2$ , then by using the refractive index from one medium to another medium formula,
$\Rightarrow {}_3{\mu ^2} = \dfrac{{{\mu _3}}}{{{\mu _2}}}\,.......................\left( 2 \right)$
Similarly, for ${}_4{\mu ^3}$, the refractive index from medium $4$ to medium $3$ , then by using the refractive index from one medium to another medium formula,
$\Rightarrow {}_4{\mu ^3} = \dfrac{{{\mu _4}}}{{{\mu _3}}}\,.......................\left( 3 \right)$
Now, the product of ${}_2{\mu ^1} \times {}_3{\mu ^2} \times {}_4{\mu ^3}$ is given as the product of the equation (1), equation (2) and equation (3), then
$\Rightarrow {}_2{\mu ^1} \times {}_3{\mu ^2} \times {}_4{\mu ^3} = \dfrac{{{\mu _2}}}{{{\mu _1}}}\, \times \dfrac{{{\mu _3}}}{{{\mu _2}}} \times \dfrac{{{\mu _4}}}{{{\mu _3}}}$
Now cancelling the same terms ${\mu _2}$ and ${\mu _3}$ from numerator and the denominator in the RHS, then
${}_2{\mu ^1} \times {}_3{\mu ^2} \times {}_4{\mu ^3} = \dfrac{{{\mu _4}}}{{{\mu _1}}}\, = {}_4{\mu ^1}$
Thus, the above equation shows the product of ${}_2{\mu ^1} \times {}_3{\mu ^2} \times {}_4{\mu ^3}$.

Hence, option (C) is the correct answer.

Note:
The final answer shows the refractive index of light goes from medium $4$ to medium $1$. In physics, refraction is the change in the direction of the wave that falls from one medium to another or from a gradual change in the medium. Refraction of the light is the most commonly observed phenomenon, but other waves such as sound waves and water waves also show refraction.