Question
What is lateral displacement? Show that a ray incident obliquely on a rectangular glass slab, emerges parallel to the incident ray
✓
Answer
$\rightarrow$ Place a glass slab on a plane sheet of paper fixed on a drawing board.
$\rightarrow$ Draw its boundary.
$\rightarrow$ Fix two pins A and B on one side of the slab.
$\rightarrow$ Look through the opposite side of the slab and fix two pins C and D in such a way that all the four pins A, B, C and D appear to be in a straight line.
$\rightarrow$ CD is the emergent ray. Let it subtend an angle e with the normal.
$\rightarrow$ In the first refraction from medium a (air) to medium b (glass), i is the angle of incidence and r is the angle of refraction.
$n_{ ba }=\frac{\sin t}{\sin r}\ldots\ldots (1)$
In the second refraction at E, the angle of incidence in medium bis r and the angle of refraction in medium a is e.
$ n_{ab}=\frac{\sin r}{\sin e} \ldots\ldots (2)$
From (1) and (2), we have
$ n_{ba} \times n_{ab}=\frac{\sin i}{\sin r} \times \frac{\sin r}{\sin e} $
But $n_{ ba } \times n_{ ab }=1$, according to the principle of reversibility.
$\begin{array}{l}\text { Hence, } \quad 1=\frac{\sin t}{\sin e} \\ \therefore \sin i=\sin e \\ \therefore i=e\end{array}$
Thus, the angle of emergence is equal to the angle of incidence. Therefore, incident ray produced (or extended), i.e., AOd is parallel to emergent ray CD.
Thus, the emergent ray is parallel to the incident ray, but displaced sideways as shown in the figure. This sideways displacement FE is known as lateral displacement.
$\rightarrow$ Draw its boundary.
$\rightarrow$ Fix two pins A and B on one side of the slab.
$\rightarrow$ Look through the opposite side of the slab and fix two pins C and D in such a way that all the four pins A, B, C and D appear to be in a straight line.
$\rightarrow$ CD is the emergent ray. Let it subtend an angle e with the normal.
$\rightarrow$ In the first refraction from medium a (air) to medium b (glass), i is the angle of incidence and r is the angle of refraction.
$n_{ ba }=\frac{\sin t}{\sin r}\ldots\ldots (1)$
In the second refraction at E, the angle of incidence in medium bis r and the angle of refraction in medium a is e.
$ n_{ab}=\frac{\sin r}{\sin e} \ldots\ldots (2)$
From (1) and (2), we have
$ n_{ba} \times n_{ab}=\frac{\sin i}{\sin r} \times \frac{\sin r}{\sin e} $
But $n_{ ba } \times n_{ ab }=1$, according to the principle of reversibility.
$\begin{array}{l}\text { Hence, } \quad 1=\frac{\sin t}{\sin e} \\ \therefore \sin i=\sin e \\ \therefore i=e\end{array}$
Thus, the angle of emergence is equal to the angle of incidence. Therefore, incident ray produced (or extended), i.e., AOd is parallel to emergent ray CD.
Thus, the emergent ray is parallel to the incident ray, but displaced sideways as shown in the figure. This sideways displacement FE is known as lateral displacement.
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