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Question: If the refractive index of water and glass with respect to air are respectively \(\dfrac{4}{3}\) and...

If the refractive index of water and glass with respect to air are respectively 43\dfrac{4}{3} and 32\dfrac{3}{2} then what will be the refractive index of glass with respect to water.

& A.~~~~~~~\dfrac{9}{8}~ \\\ & B.~~~~~~~\dfrac{8}{9}~ \\\ & C.~~~~~~~\dfrac{4}{3}~ \\\ & D.~~~~~~~\dfrac{3}{2}~ \\\ \end{aligned}$$
Explanation

Solution

Hint: Refractive index of A with respect to B is the ratio of refractive index of A to refractive index of B. We are given a refractive index of water with respect to air and refractive index of glass with respect to air. So, to find the refractive index of glass with respect to water we can simply divide the refractive index of water with respect to air and refractive index of glass with respect to air.

Complete Step By Step Solution:
Refractive index of any substance is the ratio of speed of light in vacuum to speed of light in that substance. Mathematically,
μ=c0cs\mu =\dfrac{{{c}_{0}}}{{{c}_{s}}} , where
c0{{c}_{0}}is speed of light in vacuum and cs{{c}_{s}}is speed of light in that substance
Refractive index of medium A with respect to medium B can mathematically be written BμA=μAμB{}^{B}{{\mu }_{A}}=\dfrac{{{\mu }_{A}}}{{{\mu }_{B}}}, where
μA{{\mu }_{A}} is refractive index of medium A and μB{{\mu }_{B}}is refractive index of medium B
We are given Refractive index of water with respect to air
μwμa=43\dfrac{{{\mu }_{w}}}{{{\mu }_{a}}}=\dfrac{4}{3} … (1)
whereμw{{\mu }_{w}} is refractive index of water and μa{{\mu }_{a}} is refractive index of air
Also, we are given
μgμa=32\dfrac{{{\mu }_{g}}}{{{\mu }_{a}}}=\dfrac{3}{2} … (2)
whereμg{{\mu }_{g}} is refractive index of glass and μa{{\mu }_{a}} is refractive index of air
Now, we have to obtain a refractive index of glass with respect to water. So. we have to find
μgμw\dfrac{{{\mu }_{g}}}{{{\mu }_{w}}}
On dividing equation 1 and 2
(μwμa)(μgμa)=(43)(32)\dfrac{\left( \dfrac{{{\mu }_{w}}}{{{\mu }_{a}}} \right)}{\left( \dfrac{{{\mu }_{g}}}{{{\mu }_{a}}} \right)}=\dfrac{\left( \dfrac{4}{3} \right)}{\left( \dfrac{3}{2} \right)}
Taking reciprocal of the above equation, we get-
μgμw=98\dfrac{{{\mu }_{g}}}{{{\mu }_{w}}}=\dfrac{9}{8}
Hence, the correct option is A.

Note: This question can also be done by converting the refractive index as ratio of speed of light in vacuum and speed of light in medium. But both of the methods are quite similar. Also, we should remember that the refractive index of any medium is always greater than 1. This is because the refractive index of any substance is the ratio of speed of light in vacuum to speed of light in that substance, and speed of light is highest in vacuum. Hence, the denominator is always smaller than the numerator. So, the refractive index is always greater than 1.