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

Answer 1

Answer: (C)

Answer 2

:

From figure, in right angled $\Delta \mathrm { CDB }$

$\therefore \sin ( \mathrm { i } - \mathrm { r } ) = \frac { \mathrm { CD } } { \mathrm { BC } } = \frac { \mathrm { d } } { \mathrm { BC } }$

or d = BC sin (i – r) …(i)

Also, in right angled $\Delta \mathrm { CNB }$

$\cos \mathrm { r } = \frac { \mathrm { BN } } { \mathrm { BC } } = \frac { \mathrm { t } } { \mathrm { BC } }$

$\mathrm { BC } = \frac { \mathrm { t } } { \cos \mathrm { r } }$ …(ii)

Substitute equation (ii) in equation (i), we get

$\mathrm { d } = \frac { \mathrm { t } } { \cos \mathrm { r } } \sin ( \mathrm { i } - \mathrm { r } )$

For small angles sin $( \mathrm { i } - \mathrm { r } ) \approx \mathrm { i } - \mathrm { r } ; \cos \mathrm { r } \approx 1$

$\mathrm { d } = \mathrm { t } ( \mathrm { i } - \mathrm { r } ) , \mathrm { d } = \mathrm { it } \left[ 1 - \frac { \mathrm { r } } { \mathrm { i } } \right]$