Question

For a ray of light refracting from medium $1$ to medium $2$ with plane interface, if unit vectors in the direction of incident ray, normal and the refracted ray are represented by ${\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{k} _1}$,$\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{n} $and${\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{k} _2}$respectively, and the angles of incidence and refraction by${\theta _1}$and ${\theta _2}$respectively, then the Snell’s law for refraction at the interface can be written in vector form as

A.${\mu _2}{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{k} _2} - {\mu _1}{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{k} _1} = \left( {{\mu _1}\cos {\theta _1} - {\mu _2}\cos {\theta _2}} \right)\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{n} $
B.${\mu _2}{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{k} _2} - {\mu _1}{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{k} _1} = \left( {{\mu _2}\sin {\theta _2} - {\mu _1}\sin {\theta _1}} \right)\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{n} $
C.${\mu _1}{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{k} _2} - {\mu _2}{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{k} _1} = \left( {{\mu _2}\cos {\theta _2} - {\mu _1}\cos {\theta _1}} \right)\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{n} $
D.${\mu _1}{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{k} _1} - {\mu _2}{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{k} _2} = \left( {{\mu _2}\sin {\theta _2} - {\mu _1}\sin {\theta _1}} \right)\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{n} $

Answer 1

Hint: To answer this question, we first need to write the Snell’s law for refraction in the form of a vector cross product. Then, manipulating the expression with the help of the vector triple product, we will get the final result.
Formula Used:
${\mu _1}\sin {\theta _1} = {\mu _2}\sin {\theta _2}$, here ${\mu _1}$and\[{\mu _2}\]are the refractive indices of the medium$1$ and the medium$2$respectively, and \[{\theta _1}\]and \[{\theta _2}\]are the angles made by the ray of light with the normal in the respective mediums.

Complete answer:
The situation is shown in the figure below.

Applying Snell’s law at the interface, we have
${\mu _1}\sin {\theta _1} = {\mu _2}\sin {\theta _2}$
We know from trigonometry, that $\sin \theta = \sin \left( {{{180}^ \circ } - \theta } \right)$
So, the Snell’s law can also be written as
${\mu _1}\sin \left( {{{180}^ \circ } - {\theta _1}} \right) = {\mu _2}\sin \left( {{{180}^ \circ } - {\theta _2}} \right)$ (1)
Now, we can see from the above figure that the angle between the unit vectors ${\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{k} _1}$and $\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{n} $ is equal to${180^ \circ } - {\theta _1}$
So, we have \[{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{k} _1} \times \overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{n} = 1 \times 1 \times \sin \left( {{{180}^ \circ } - {\theta _1}} \right)\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{c} \]
\[{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{k} _1} \times \overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{n} = \sin \left( {{{180}^ \circ } - {\theta _1}} \right)\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{c} \] (2)
Similarly, we have
\[{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{k} _2} \times \overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{n} = \sin \left( {{{180}^ \circ } - {\theta _2}} \right)\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{c} \] (3)
Here \[\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{c} \]is the unit vector perpendicular to the plane of the paper.
Substituting (2) and (3) in (1), we get
\[{\mu _1}\left( {{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{k} }_1} \times \overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{n} } \right) = {\mu _2}\left( {{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{k} }_2} \times \overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{n} } \right)\]
Now, we take the cross product of the unit vector $\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{n} $ with both the sides
\[{\mu _1}\left( {\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{n} \times \left( {{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{k} }_1} \times \overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{n} } \right)} \right) = {\mu _2}\left( {\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{n} \times \left( {{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{k} }_2} \times \overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{n} } \right)} \right)\]
Writing the values of the vector triple products on both sides
\[{\mu _1}\left( {\left( {\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{n} \cdot \overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{n} } \right){{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{k} }_1} - \left( {\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{n} \cdot {{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{k} }_1}} \right)\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{n} } \right) = {\mu _2}\left( {\left( {\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{n} \cdot \overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{n} } \right){{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{k} }_2} - \left( {\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{n} \cdot {{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{k} }_2}} \right)\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{n} } \right)\] (4)
We have \[\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{n} \cdot \overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{n} = 1 \times 1 = 1\]
Also,
\[\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{n} \cdot {\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{k} _1} = 1 \times 1 \times \cos \left( {{{180}^ \circ } - {\theta _1}} \right)\]
\[\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{n} \cdot {\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{k} _1} = - \cos {\theta _1}\]
Similarly
\[\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{n} \cdot {\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{k} _2} = - \cos {\theta _2}\]
Putting these values in (4), we get
\[{\mu _1}\left( {{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{k} }_1} - \left( { - \cos {\theta _1}} \right)\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{n} } \right) = {\mu _2}\left( {{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{k} }_2} - \left( { - \cos {\theta _2}} \right)\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{n} } \right)\]
\[{\mu _1}\left( {{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{k} }_1} + \left( {\cos {\theta _1}} \right)\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{n} } \right) = {\mu _2}\left( {{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{k} }_2} + \left( {\cos {\theta _2}} \right)\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{n} } \right)\]
On multiplying, we get
\[{\mu _1}{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{k} _1} + {\mu _1}\left( {\cos {\theta _1}} \right)\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{n} = {\mu _2}{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{k} _2} + {\mu _2}\left( {\cos {\theta _2}} \right)\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{n} \]
On rearranging the terms, we get
\[{\mu _1}\cos {\theta _1}\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{n} - {\mu _2}\cos {\theta _2}\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{n} = {\mu _2}{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{k} _2} - {\mu _1}{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{k} _1}\]
Taking $\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{n} $as common, we finally get
\[{\mu _2}{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{k} _2} - {\mu _1}{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{k} _1} = \left( {{\mu _1}\cos {\theta _1} - {\mu _2}\cos {\theta _2}} \right)\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{n} \]

Hence, the correct answer is option A.

Note:
We used the vector triple product in this question just to manipulate our answer in the same form as the options were given. The idea is simple, as the vector triple product involves subtraction of the vectors, the form in which the options are given, so we used this method.