Question

A ray of light is incident on a thick slab of glass of thickness $t$ as shown in the figure. The emergent ray is parallel to the incident ray but displaced sideways by a distance $d$. If the angles are small then $d$ is,

• A. $t\left(1-\frac{i}{r}\right)$
• B. $r t\left(1-\frac{i}{r}\right)$
• C. it $\left(1-\frac{r}{i}\right)$
• D. $t\left(1-\frac{r}{i}\right)$

Answer 1

Answer: (C)

From figure, in right angled $\Delta CDB$
$\angle CBD=(i-r)$
$\therefore \sin (i-r)=\frac{C D}{B C}=\frac{d}{BC}$
or $ d=B C \sin (i-r) \,\,\,\,\dots(i)$
Also, in right angled $\Delta CNB$
$\cos r=\frac{B N}{B C}=\frac{t}{BC}$ or
$BC=\frac{t}{\cos r}\,\,\,\,\dots(ii)$
Now, from equation (i), we get $d=\frac{t}{\cos r} \sin (i-r)$
For small angles $\sin (i-r) \approx i-r ; \cos r \approx 1$
$d=t(i-r), d=i t\left(1-\frac{r}{i}\right)$