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Question: The critical angle for refraction from glass to air is \[30^\circ \] and that from water to air is \...

The critical angle for refraction from glass to air is 30∘30^\circ and that from water to air is 37∘37^\circ . Find the critical angle for refraction from glass to water.

Explanation

Solution

We will first use Snell’s law, i.e., n1sin⁡i=n2sin⁡r{n_1}\sin i = {n_2}\sin r. And then use the formula for determining the critical angle, θC=sin⁡−1(μ2μ1){\theta _C} = {\sin ^{ - 1}}\left( {\dfrac{{{\mu _2}}}{{{\mu _1}}}} \right)

Complete step by step answer:
On considering the first situation i.e., when the light ray passes from glass to air and θ=30∘\theta = 30^\circ .
Here, we will apply Snell’s law,
n1sin⁡i=n2sin⁡r{n_1}\sin i = {n_2}\sin r
Where, n1={n_1} = refractive index of glass; n2={n_2} = refractive index of air i.e., one; angle of incidence is 30∘30^\circ and angle of refraction is 90∘90^\circ .
nGsin⁡30∘=1×sin⁡90∘{n_G}\sin 30^\circ = 1 \times \sin 90^\circ
⇒nG×sin⁡30∘=1\Rightarrow {n_G} \times \sin 30^\circ = 1 ………………………(i.)
Now, let’s consider the second situation where light rays pass from water to air and θ=37∘\theta = 37^\circ .
nWsin⁡37∘=1×sin⁡90∘{n_W}\sin 37^\circ = 1 \times \sin 90^\circ
⇒nW×sin⁡37∘=1\Rightarrow {n_W} \times \sin 37^\circ = 1 ……………………. (ii.)
We know the formula for finding the critical angle, i.e., θC=sin⁡−1(μ2μ1){\theta _C} = {\sin ^{ - 1}}\left( {\dfrac{{{\mu _2}}}{{{\mu _1}}}} \right)
⇒θC=sin⁡−1(nWnG)\Rightarrow {\theta _C} = {\sin ^{ - 1}}\left( {\dfrac{{{n_W}}}{{{n_G}}}} \right) ………………………….. (iii.)
On dividing eq. (ii.) with eq. (i.)
nWsin⁡37∘nGsin⁡30∘=1\dfrac{{{n_W}\sin 37^\circ }}{{{n_G}\sin 30^\circ }} = 1
nWnG=sin⁡30∘sin⁡37∘\dfrac{{{n_W}}}{{{n_G}}} = \dfrac{{\sin 30^\circ }}{{\sin 37^\circ }}
Now, substitute this value in eq., (iii.),
⇒θC=sin⁡−1(sin⁡30∘sin⁡37∘)\Rightarrow {\theta _C} = {\sin ^{ - 1}}\left( {\dfrac{{\sin 30^\circ }}{{\sin 37^\circ }}} \right)
We know the value of sin⁡37∘\sin 37^\circ and sin⁡30∘\sin 30^\circ . On putting those values in the above relation, we will get,
θC=sin⁡−1(56){\theta _C} = {\sin ^{ - 1}}\left( {\dfrac{5}{6}} \right)
This is our required angle.

Note: There are other alternatives to find the final answer.
We can directly substitute the values in equations (i.) and (ii.) and get the values of the refractive index of glass and water.
We can use the critical angle formula and put the already known values of the refractive indices of glass and water.