Question

The wavelength of the light in two liquid’s $x$ and $y$ is $3500\;\mathop {\rm{A}}\limits^{\rm{0}}$ and $7000\;\mathop {\rm{A}}\limits^{\rm{0}}$. Then the critical angle of $x$ relative to $y$ will be
(A)$60^\circ$
(B)$45^\circ$
(C) $30^\circ$
(D)$15^\circ$

Answer 1

If a wave travels in a straight line, then its velocity can be calculated by division of velocity (V) by frequency (f) of the wave. The mathematical expression for wavelength ($\lambda$) is given as follows,
$\lambda = \dfrac{V}{f}$

Complete step-by-step solution:
The wavelength of light in the liquid $x$ is :${\lambda _x} = 3500\;\mathop {\rm{A}}\limits^{\rm{0}}$
The wavelength of light in the liquid $y$ is :${\lambda _y} = 7000\;\mathop {\rm{A}}\limits^{\rm{0}}$
We know that the refractive index of a liquid is the division of the velocity of the light by the velocity of the wave in that liquid.
The expression for the relation between the refractive index and the velocity is given as follows,
$n = \dfrac{C}{V}$
Here, $C$ is the velocity of the light, $n$ refractive index of the medium and $V$ is the velocity of the wave in the medium.
Now we will calculate the refractive index for both the liquid.
$\begin{array}{l} {n_x} = \dfrac{C}{{{V_x}}}............{\rm{(1)}}\\\ {n_y} = \dfrac{C}{{{V_y}}}............{\rm{(2)}} \end{array}$
Here, ${n_x}$ and ${n_y}$ is the refractive index of the given liquid $x$ and $y$ respectively and ${V_x}$, ${V_y}$ is the velocity of light in liquid $x$ and liquid $y$ respectively.
Divide the equation (1) by the equation (2).
$\dfrac{{{n_x}}}{{{n_y}}} = \dfrac{{{V_y}}}{{{V_x}}}...................{\rm{(3)}}$
The relation between the wavelength and the velocity of light is given as follows,
$V = f\lambda$
Now we will calculate the velocity of light in both the liquid.

{V_x} = f{\lambda _x}\\\ {V_y} = f{\lambda _y} \end{array}$$ Here, $f$ is the frequency of the light wave and $${\lambda _x}$$,$${\lambda _y}$$is the wavelength of light in liquid $x$ and liquid $y$ respectively. Substitute all the values in the equation (3). $\begin{array}{l} \dfrac{{{n_x}}}{{{n_y}}} = \dfrac{{f{\lambda _y}}}{{f{\lambda _x}}}\\\ \dfrac{{{n_x}}}{{{n_y}}} = \dfrac{{{\lambda _y}}}{{{\lambda _x}}}...............{\rm{(4)}} \end{array}$ we know that critical angle of medium $x$ relative to $y$ is, $\begin{array}{l} \theta = {\sin ^{ - 1}}\left( {\dfrac{1}{{\dfrac{{{n_x}}}{{{n_y}}}}}} \right)\\\ \Rightarrow \theta = {\sin ^{ - 1}}\left( {\dfrac{1}{{\dfrac{{{\lambda _y}}}{{{\lambda _x}}}}}} \right) \end{array}$ Substitute all the values in the above expression, $\begin{array}{l} \Rightarrow \theta = {\sin ^{ - 1}}\left( {\dfrac{1}{{\dfrac{{7000}}{{3500}}}}} \right)\\\ \Rightarrow \theta = {\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right)\\\ \Rightarrow \theta = 30^\circ \end{array}$ Therefore, the critical angle of $x$ relative to $y$ will be $30^\circ $ and **the correct answer is option (C).** **Note:** Wavelength of the electromagnetic wave is the distance covered by the specific particle in a specific time interval or the distance of the two adjacent crests or troughs. The wave repeats its shape about these adjacent points.