A light ray strikes a 2 cm thick plate ( μ = 2.0 ) at an angle of...

Using Snell's law at the first interface:

$\sin i = \mu \sin r$

$\sin 45^\circ = 2.0 \sin r$

$\dfrac{1}{\sqrt{2}} = 2 \sin r \Rightarrow \sin r = \dfrac{1}{2\sqrt{2}}$

From this, we find $\cos r$:

$\cos r = \sqrt{1 - \sin^2 r} = \sqrt{1 - \dfrac{1}{8}} = \dfrac{\sqrt{7}}{2\sqrt{2}}$

The lateral shift $x$ of the emergent ray is given by:

$x = \dfrac{t \sin(i - r)}{\cos r} = t (\sin i - \cos i \tan r)$

Substituting the values:

x = \sqrt{2} \left( \sin 45^\circ - \cos 45^\circ\dfrac{\sin r}{\cos r} \right)

$x =$\sqrt{2} \left( \dfrac{1}{\sqrt{2}} - \dfrac{1}{\sqrt{2}} \dfrac{1 / (2\sqrt{2})}{\sqrt{7} / (2\sqrt{2})} \right)$<br><br>$x = \sqrt{2} \left( \dfrac{1}{\sqrt{2}} - \dfrac{1}{\sqrt{2}} \dfrac{1}{\sqrt{7}} \right)<br><br>x = 1 - $\dfrac{1}{\sqrt{7}}<br><br>Using \sqrt{7} \approx$2.645$:<br><br>$x = 1 - $\dfrac{1}{2.645} \approx$1 - 0.378 = 0.622 \text{ cm}