Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • A ray of light travelling along the line after striking a plane mi

A ray of light travelling along the line \mathrm{2 x-3 y+5=0} after striking a plane mirror lying along the line\mathrm{ x+y=2} gets reflected. Find the equation of the straight line containing the reflected ray.

Option: 1

\mathrm{2 x-3 y+3=0}
 


Option: 2

\mathrm{3 x-2 y+3=0}


Option: 3

\mathrm{3 x+2 y-3=0}
 


Option: 4

\mathrm{2 y=3 x-2}


Answers (1)

best_answer

The point of intersection of the lines \mathrm{2 x-3 y+5=0 \: and \: x+y=2} is \mathrm{\left(\frac{1}{5}, \frac{9}{5}\right)\left(\frac{1}{5}, \frac{9}{5}\right)} is the point of incidence.
Slope \mathrm{ \mathrm{m} } of the normal to the mirror (i.e. normal to the line \mathrm{x+y=2} ) is 1 .

Now the incident ray and reflected ray both are equally inclined to the normal and are on opposite side of it.

Slope of incident ray \mathrm{m_1=2 / 3}

Let the slope of the reflected ray be \mathrm{=m_2}

Then

\mathrm{ \frac{m_1-m}{1+m_1 m}=\frac{m-m_2}{1+m_2 m} }

i.e.\mathrm{ \frac{\frac{2}{3}-1}{1+\frac{2}{3} \times 1}=\frac{1-m_2}{1+m_2 \times 1} }

\mathrm{ \therefore m_2=3 / 2, \quad \therefore } the equation of the straight line containing the reflected ray is

\mathrm{ y-\frac{9}{5}=\frac{3}{2}\left(x-\frac{1}{5}\right) \quad \text { i.e. } \quad 3 x-2 y+3=0 }

Hence option 2 is correct.
 

Posted by

rishi.raj

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Latest Question