3.1 Shift due to a glass slab

$OA = x$

${I_1}A = \mu x$ (Apparent image due to refraction from the 1st surface)

$$\begin{equation} \begin{aligned} {I_1}B = {I_1}A + AB \\ {I_1}B = \mu x + t \\\end{aligned} \end{equation} $$

$IB = \frac{{{I_1}B}}{\mu }$ (Apparent image due to reflection from the 2nd surface).

$$\begin{equation} \begin{aligned} IB = \left( {\frac{{\mu x + t}}{\mu }} \right) \\ IB = \left( {x + \frac{t}{\mu }} \right) \\\end{aligned} \end{equation} $$

Shift $(OI)$ is given by,

$$\begin{equation} \begin{aligned} OI = (OA + AB) - (BI) \\ OI = (x + t) - \left( {x + \frac{t}{\mu }} \right) \\ OI = t\left[ {1 - \frac{t}{\mu }} \right] \\\end{aligned} \end{equation} $$

When the medium is same on both sides of a glass slab, then the deviation of the emergent ray is zero. That is the emergent ray is parallel to the incident ray but it does suffer lateral shift or lateral displacement with respect to the incident ray as shown in the figure.

Mathermatically lateral shift is given by, $$d = \left( {1 - \frac{1}{\mu }} \right)ti$$

$$AB\cos r = AC$$ or $$AB = \frac{t}{{\cos r}}$$

$$d = AB\sin (i - r)$$ So, $$\begin{equation} \begin{aligned} d = \frac{t}{{\cos r}}\left[ {\sin i\cos r - \cos i\sin r} \right] \\ d = t\left[ {\sin i - \cos i\tan r} \right]\quad ...(i) \\\end{aligned} \end{equation} $$

From Snell's law,

$$\frac{{\sin i}}{{\sin r}} = \mu $$ or $$\sin r = \frac{{\sin i}}{\mu }$$

Therefore, $$\tan r = \frac{{\sin r}}{{\cos r}} = \frac{{\sin r}}{{\sqrt {1 - {{\sin }^2}r} }}$$

$$\tan r = \frac{{\frac{{\sin i}}{\mu }}}{{\sqrt {1 - {{\left( {\frac{{\sin i}}{\mu }} \right)}^2}} }} = \frac{{\sin i}}{{\sqrt {{\mu ^2} - {{\sin }^2}i} }}\quad ...(i)$$

From equation $(i)$ and $(ii)$ we get,

$$d = t\left[ {\sin i - \frac{{\sin i\cos i}}{{\sqrt {{\mu ^2} - {{\sin }^2}i} }}} \right]$$

For $i \to 0$,

$$d = \mathop {\lim }\limits_{i \to 0} \left[ {1 - \frac{{\cos i}}{{\sqrt {{\mu ^2} - {{\sin }^2}i} }}} \right]t\sin i$$