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A ray of light is sent along the line 3x + y - 7 = 0. Upon reaching the line 3x - 2y + 5  = 0, the ray is reflected from it. Then the equation of the line containing the reflected ray is

Option: 1

3x + 41y + 161 = 0


Option: 2

3x - 39y + 161 = 0


Option: 3

3x - 41y + 161 = 0


Option: 4

3x + 39y + 161 = 0


Answers (1)

best_answer

 

 

Reflection of Light -

Reflection of Light

Laws of reflection

  1. The incident ray, the normal ray and the reflected ray to a surface at the point of an incident all lie on the same plane.

  2. The angle of incident = angle of reflection

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As from the figure,

To get coordinate of P, solve the equation of line together   

3x + y - 7 = 0

3x - 2y + 5  = 0

We get, P = (1, 4)

Let, the slope of reflected ray is m

The slope of mirror line is 3/2 

Then the slope of the line perpendicular to mirror line is (i.e.  PN) -2/3

and slope of the incident ray is -3

Line PN is equally inclined to the line PR and IP

 

\\\mathrm{\left ( \frac{(-3)-\left ( -\frac{2}{3} \right )}{1+(-3)\left ( -\frac{2}{3} \right )} \right )}=-\left ( \frac{m-\left ( -\frac{2}{3} \right )}{1+m\left ( -\frac{2}{3} \right )} \right )\\\mathrm{m=\frac{3}{41}}\\\mathrm{Equation\;of\;reflected\;ray\;is}\\\mathrm{y-4=\frac{3}{41}(x-1)}

3x - 41y + 161 = 0

Altair

IMAGE METHOD

Choose A(0, 7) any point on reflected ray i.e. IP

And let  B(α, β) is the image of point A about the mirror line 3x - 2y + 5  = 0, then

 

\\\frac{\alpha-0}{3}=\frac{\beta-7}{-2}=-2\left ( \frac{3(0)-2(7)+5}{3^2+(-2)^2} \right )\\\alpha=\frac{54}{13}\\\beta=\frac{55}{13}\\Now,equation\;of\;line\;passing\;through\;(1,4)\;\;and\;\;\left ( \frac{54}{13},\frac{55}{13} \right )\\y-4=\frac{\frac{55}{13}-4}{\frac{54}{13}-1}\left(x-1\right)\\3\left(x-1\right)=41y-164\\3x-41y+161=0

Posted by

Devendra Khairwa

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