1. Tardigrade
  2. Question
  3. Physics
  4. A ray is incident on a plane surface. If hati + hatj - hatk represents a vector along the direction of incident ray. hati + hatj is a vector along normal on incident point in the plane of incident and reflected ray, then vector along the direction of reflected ray is

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. A ray is incident on a plane surface. If $\hat{i} + \hat{j} - \hat{k}$ represents a vector along the direction of incident ray.$\hat{i} + \hat{j} $ is a vector along normal on incident point in the plane of incident and reflected ray, then vector along the direction of reflected ray is

JIPMERJIPMER 2011Ray Optics and Optical Instruments

Solution:

image
According to law of reflection in vector form,
$\hat{n_2} = \hat{n_1} - 2 (\hat{n_1} .\hat{n}) \hat{n}$
Here, $\hat{n_1}$ = the unit vector along incident ray
$= \frac{\hat{i} + \hat{j} - \hat{k}}{\sqrt{3}} $
$\hat{n}$ = unit vector along normal on incident point
$= \frac{\hat{i} + \hat{j}}{\sqrt{2}}$
$\hat{n_2}$ =unit vector along the direction of reflected ray
Using the formula, we get
$\hat{n_2} = \frac{\hat{i} + \hat{j} - \hat{k}}{\sqrt{3}} - 2 \bigg\{ \bigg( \frac{\hat{i} + \hat{j} - \hat{k}}{\sqrt{3}}\bigg) . \bigg( \frac{\hat{i} + \hat{j}}{\sqrt{2}} \bigg). \bigg( \frac{\hat{i} + \hat{j} }{\sqrt{2}} \bigg) \bigg\}$
$= \frac{\hat{i} + \hat{j} - \hat{k}}{\sqrt{3}} - 2 \bigg\{ \frac{1}{\sqrt{6}} + \frac{1}{\sqrt{6}} \bigg\} \frac{\hat{i} + \hat{j}}{\sqrt{2}}$
$= \frac{\hat{i} + \hat{j} - \hat{k}}{\sqrt{3}} - \frac{4}{\sqrt{6}} \bigg( \frac{\hat{i} + \hat{j}}{\sqrt{2}} \bigg)$
$= \frac{\hat{i} + \hat{j} - \hat{k}}{\sqrt{3}} - \frac{4\hat{i}}{\sqrt{12}}- \frac{4\hat{j}}{\sqrt{12}}$
$= \frac{2 \hat{i} +2 \hat{j} - 2\hat{k} -4\hat{i} - 4\hat{j} }{2\sqrt{3}} $ $(\because \, \sqrt{12} = 2 \sqrt{3} )$
$= \frac{-2 \hat{i} -2 \hat{j} - 2\hat{k} }{2\sqrt{3}} = \frac{- \hat{i} - \hat{j} -\hat{k}}{\sqrt{3}}$
$= - \frac{1}{\sqrt{3}} ( \hat{i} + \hat{j} +\hat{k})$