Questions & Answers

Question

Answers

Answer

Verified

155.7k+ views

In a concave mirror if the ray of light is parallel to the axis of the mirror, then the reflected ray will pass through the focus of the concave mirror.

As per the information given, we can draw the diagram as follow.

Let the ray of light meet the concave mirror at point P. Then Pf is the reflected ray where f is the focus of the concave mirror.

The equation of parabola or the equation of the concave mirror is,

${{y}^{2}}=4ax$

So, the coordinate of the point f will be $\left( a,0 \right)$.

As it is given that ray of light with equation $y=b$ strikes the concave mirror, so substituting the value of ‘y’ we get,

${{b}^{2}}=4ax$

$x=\dfrac{{{b}^{2}}}{4a}$

So, the coordinates of the point P is $\left( \dfrac{{{b}^{2}}}{4a},b \right)$ .

Now the slope of the line passing through two points $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$is given by,

$m=\dfrac{\left( {{y}_{2}}-{{y}_{1}} \right)}{\left( {{x}_{2}}-{{x}_{1}} \right)}$

So, the slope of the line passing through point \[f\left( a,0 \right)\] and \[P\left( \dfrac{{{b}^{2}}}{4a},b \right)\] is,

$m=\dfrac{b-0}{\dfrac{{{b}^{2}}}{4a}-a}$

$\Rightarrow m=\dfrac{b}{\dfrac{{{b}^{2}}-4{{a}^{2}}}{4a}}$

$\Rightarrow m=\dfrac{4ab}{{{b}^{2}}-4{{a}^{2}}}$

Now we know the line of equation passing through the point $\left( {{x}_{1}},{{y}_{1}} \right)$is,

$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$

where ‘m’ is the slope of the line.

So, the equation of line passing through the point \[f\left( a,0 \right)\]with slope, $m=\dfrac{4ab}{{{b}^{2}}-4{{a}^{2}}}$, is,

$y-0=\left( \dfrac{4ab}{{{b}^{2}}-4{{a}^{2}}} \right)\left( x-a \right)$

$\Rightarrow \left( {{b}^{2}}-4{{a}^{2}} \right)y=4ab\left( x-a \right)$

$\Rightarrow 4abx-4{{a}^{2}}b-\left( {{b}^{2}}-4{{a}^{2}} \right)y=0$

$\Rightarrow 4abx-\left( {{b}^{2}}-4{{a}^{2}} \right)y-4{{a}^{2}}b=0$

Now as the reflected ray is passing through the focus, i.e., f, so the equation of reflected ray is,

$4abx-\left( {{b}^{2}}-4{{a}^{2}} \right)y-4{{a}^{2}}b=0$

Note:- In this particular type of questions first find the intersection point of line and parabola, then find the slope of line then apply equation of line formula you will get your answer

As per the information given, we can draw the diagram as follow.

Let the ray of light meet the concave mirror at point P. Then Pf is the reflected ray where f is the focus of the concave mirror.

The equation of parabola or the equation of the concave mirror is,

${{y}^{2}}=4ax$

So, the coordinate of the point f will be $\left( a,0 \right)$.

As it is given that ray of light with equation $y=b$ strikes the concave mirror, so substituting the value of ‘y’ we get,

${{b}^{2}}=4ax$

$x=\dfrac{{{b}^{2}}}{4a}$

So, the coordinates of the point P is $\left( \dfrac{{{b}^{2}}}{4a},b \right)$ .

Now the slope of the line passing through two points $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$is given by,

$m=\dfrac{\left( {{y}_{2}}-{{y}_{1}} \right)}{\left( {{x}_{2}}-{{x}_{1}} \right)}$

So, the slope of the line passing through point \[f\left( a,0 \right)\] and \[P\left( \dfrac{{{b}^{2}}}{4a},b \right)\] is,

$m=\dfrac{b-0}{\dfrac{{{b}^{2}}}{4a}-a}$

$\Rightarrow m=\dfrac{b}{\dfrac{{{b}^{2}}-4{{a}^{2}}}{4a}}$

$\Rightarrow m=\dfrac{4ab}{{{b}^{2}}-4{{a}^{2}}}$

Now we know the line of equation passing through the point $\left( {{x}_{1}},{{y}_{1}} \right)$is,

$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$

where ‘m’ is the slope of the line.

So, the equation of line passing through the point \[f\left( a,0 \right)\]with slope, $m=\dfrac{4ab}{{{b}^{2}}-4{{a}^{2}}}$, is,

$y-0=\left( \dfrac{4ab}{{{b}^{2}}-4{{a}^{2}}} \right)\left( x-a \right)$

$\Rightarrow \left( {{b}^{2}}-4{{a}^{2}} \right)y=4ab\left( x-a \right)$

$\Rightarrow 4abx-4{{a}^{2}}b-\left( {{b}^{2}}-4{{a}^{2}} \right)y=0$

$\Rightarrow 4abx-\left( {{b}^{2}}-4{{a}^{2}} \right)y-4{{a}^{2}}b=0$

Now as the reflected ray is passing through the focus, i.e., f, so the equation of reflected ray is,

$4abx-\left( {{b}^{2}}-4{{a}^{2}} \right)y-4{{a}^{2}}b=0$

Note:- In this particular type of questions first find the intersection point of line and parabola, then find the slope of line then apply equation of line formula you will get your answer