Explanation

(b) In $\mathrm{\Delta}BCE$
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$\sin{(\theta - r) = \frac{CE}{BC}}$
$CE = BC\sin(\theta - r)$
$or\ d = BC\sin(\theta - r)\ \ \ \ \ \ \ldots(i)$
In $\mathrm{\Delta}BMC$
$\cos{r = \frac{BM}{BC}}$
$or\ BC = \frac{BM}{\cos r} = \frac{t}{\cos r}\ \ \ \ \ \ \ \ \ \ldots(ii)$
From Eqs (i) and (iii) we get
$d = \frac{t}{\cos r}\sin(\theta - r)$
$d = \frac{t}{\cos r}{(sin}{\theta\cos{r - \cos{\theta\sin r)}}}$
$d = t(\sin{\theta - \cos\theta\tan r})$
If $n$ is the refractive index of material of slab (glass) w.r.t. air, then
$n = \ \frac{sin\ \theta}{sin\ r}$
For small angle
$n = \frac{\theta}{r}$
$r = \frac{\theta}{n}$
And $d = t(\theta - 1\ .r)\left\lbrack \because\sin\theta = \theta\cos{\theta = 1\ if\ \theta\ is\ small} \right\rbrack$
$d = t\left( \theta - \frac{\theta}{n} \right) = t\theta\left( 1 - \frac{1}{n} \right)$
$\therefore d = \frac{t\theta(n - 1)}{n}$