91Ó°ÊÓ

According to Snell’s law, the relation between refractive index and the angle made by the incident and refracted rays are determined by:

\({n_1}\sin {\theta _1} = {n_2}\sin {\theta _2}\)

Here,\(n\)is refractive index,\(\theta \)is the angle and 1 and 2 are for incidence and refraction.

The sound wave hits the cold air interface at an angle of \({\theta _1} = 25^\circ \).

The cold air temperature is \({T_2} = - 15^\circ \).

The warm air temperature is \({T_1} = + 15^\circ \).

The speed of sound as a function of temperature can be approximated by \(v = \left( {331 + 0.60T} \right)\;{\rm{m/s}}\), where\(T\)is in\(^\circ {\rm{C}}\).

The relation of refractive index is given by,

\(n = \frac{{\rm{c}}}{v}\)

Here, \({\rm{c}}\) is speed of light.

The relation from Snell’s law is given by,

\(\begin{aligned}{c}\frac{{\sin {\theta _2}}}{{\sin {\theta _1}}} = \frac{{{v_2}}}{{{v_1}}}\\\frac{{\sin {\theta _2}}}{{\sin {\theta _1}}} = \frac{{\left( {331 + 0.60{T_2}} \right)\;{\rm{m/s}}}}{{\left( {331 + 0.60{T_1}} \right)\;{\rm{m/s}}}}\\\sin {\theta _2} = \sin {\theta _1}\frac{{\left( {331 + 0.60{T_2}} \right)}}{{\left( {331 + 0.60{T_1}} \right)}}\end{aligned}\)

Substitute the known values in the above relation to find the angle of refraction.

\(\begin{aligned}{c}\sin {\theta _2} = \sin 25^\circ \frac{{\left( {331 + 0.60\left( { - 15} \right)} \right)}}{{\left( {331 + 0.60\left( {15} \right)} \right)}}\\{\theta _2} = \sin 25^\circ \frac{{\left( {322} \right)}}{{\left( {340} \right)}}\\{\theta _2} = {\sin ^{ - 1}}\left( {0.4002} \right)\\{\theta _2} \simeq 24^\circ \end{aligned}\)

Hence, the angle of refraction is\(24^\circ \).