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Laws of Reflection

Laws of Reflection :- All surfaces(perfectly smooth or rough) which can reflect light, obey the following two laws of reflection :-

(i) First law :- The angle of incidence (i) is equal to the angle of reflection (r).

i.e., ∠i = ∠r

(ii) Second law :- The incident ray, the reflected ray and the normal at the point of incidence, all lie in the same plane. This plane is called the plane of incidence (or plane of reflection). This condition can be expressed mathematically as :-

$\displaystyle \vec{R}.(\vec{I}\times \vec{N})=\vec{N}.(\vec{I}\times \vec{R})=\vec{I}.(\vec{N}\times \vec{R})=0$

here $\displaystyle {\vec{I}}$ , $\displaystyle {\vec{N}}$ and $\displaystyle {\vec{R}}$ are vectors of any magnitude along incident ray, the normal and the reflected ray respectively.

Note :-

The angle of incidence and the angle of reflection are the angles made by the incident ray and the reflected ray with the normal respectively.
The laws of reflection are valid for any type of reflecting surface irrespective of their shape, size and nature.
If the angle of incidence is 0°, then the angle of reflection is also 0°. In other words if a light ray is incident normally on a reflecting surface then after reflection it retraces its path.

Problem solving trick :-

If a light ray is incident along a vector $\displaystyle \left( a\hat{i}+b\hat{j}+c\hat{k} \right)$ and the normal is along y -axis $\displaystyle (i.e.,\text{ }along\text{ }\hat{j})$ then the vector along reflected ray is $\displaystyle \left( a\hat{i}-b\hat{j}+c\hat{k} \right)$ and also a unit vector along the reflected ray is $\displaystyle \left[ \frac{\left( a\hat{i}-b\hat{j}+c\hat{k} \right)}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}} \right]$ .

Explanation :-

We can understand the above concept by ball collision analogy. Suppose a ball having linear momentum $\displaystyle {\vec{p}}$ , strikes a wall elastically. Let the normal on the wall is along the x-axis.

From above figure, it is clear that after collision, the component of linear momentum of ball along normal changes. Similarly if an incident light ray is represented by a vector $\displaystyle \left[ for\text{ }example~\text{ }\left( a\hat{i}+b\hat{j}+c\hat{k} \right) \right]$ , then after reflection only the component along normal to the reflecting surface changes.

In vector form :-

$\displaystyle {{{\hat{e}}}_{r}}={{{\hat{e}}}_{i}}-2\left( {{{\hat{e}}}_{i}}.\hat{n} \right)\hat{n}$

here

$\displaystyle \begin{array}{l}{{{\hat{e}}}_{r}}=\text{ }unit\text{ }vector\text{ }along\text{ }reflected\text{ }ray\\{{{\hat{e}}}_{i}}=\text{ }unit\text{ }vector\text{ }along\text{ incident }ray\\\hat{n}=\text{ }unit\text{ }vector\text{ }along\text{ the normal}\end{array}$

Explanation of vector form :-

Example 1.

A ray of light is incident on a plane mirror along a vector $\displaystyle \left( \hat{i}+\hat{j}-\hat{k} \right)$ . The normal on incidence point is along $\displaystyle \left( \hat{i}+\hat{j} \right)$ . Find a unit vector along the reflected ray.

Solution :-

$\displaystyle {{{\hat{e}}}_{i}}=\frac{\hat{i}+\hat{j}-\hat{k}}{\sqrt{3}}$

$\displaystyle \hat{n}=\frac{\hat{i}+\hat{j}}{\sqrt{2}}$

$\displaystyle {{{\hat{e}}}_{i}}.\hat{n}=\left( \frac{\hat{i}+\hat{j}-\hat{k}}{\sqrt{3}} \right).\left( \frac{\hat{i}+\hat{j}}{\sqrt{2}} \right)=\frac{1+1}{\sqrt{6}}=\frac{2}{\sqrt{6}}=\sqrt{\frac{2}{3}}$

Finally,

$\displaystyle {{{\hat{e}}}_{r}}={{{\hat{e}}}_{i}}-2\left( {{{\hat{e}}}_{i}}.\hat{n} \right)\hat{n}=\left( \frac{\hat{i}+\hat{j}-\hat{k}}{\sqrt{3}} \right)-2\left( \sqrt{\frac{2}{3}} \right)\left( \frac{\hat{i}+\hat{j}}{\sqrt{2}} \right)$

$\displaystyle \Rightarrow {{{\hat{e}}}_{r}}=\left( \frac{\hat{i}+\hat{j}-\hat{k}}{\sqrt{3}} \right)-\left( \frac{2\hat{i}+2\hat{j}}{\sqrt{3}} \right)$

$\displaystyle \Rightarrow {{{\hat{e}}}_{r}}=-\frac{1}{\sqrt{3}}\left( \hat{i}+\hat{j}+\hat{k} \right)$

Another method(Short cut):-

Using ball collision analogy, component of incident ray along the normal gets reversed while other component remain unchanged. Hence vector along reflected ray :-

$\displaystyle {{{\vec{E}}}_{r}}=-\hat{i}-\hat{j}-\hat{k}$

And unit vector along the reflected ray :-

$\displaystyle {{{\hat{e}}}_{r}}=-\frac{1}{\sqrt{3}}\left( \hat{i}+\hat{j}+\hat{k} \right)$

Example 2.

A mirror on a wall is 7/3 m tall. The bottom of mirror is 0.50 m above the floor. A lightbulb hangs on the roof 4.00 m from the wall, 35/6 m above the floor. How long is the streak of reflected light across the floor?

Solution :-

Considering the lightbulb as a point source, here is the pictorial representation of the light rays reflecting from a mirror :-

Here we need only the two light rays that strike the edges of the mirror because all other reflected rays will fall between these two.

Now

$\displaystyle {{\theta }_{1}}={{\tan }^{-1}}\left( \frac{3}{4} \right)={{37}^{\circ }}$

And

$\displaystyle {{\theta }_{2}}={{\tan }^{-1}}\left( \frac{3+\frac{7}{3}}{4} \right)={{\tan }^{-1}}\left( \frac{4}{3} \right)={{53}^{\circ }}$

The distances to the points where the rays strike the floor are :-

$\displaystyle {{l}_{1}}=\frac{{}^{7}\!\!\diagup\!\!{}_{3}\;+0.5}{\tan {{\theta }_{1}}}=\frac{{}^{17}\!\!\diagup\!\!{}_{6}\;}{\tan {{37}^{\circ }}}=\frac{{}^{17}\!\!\diagup\!\!{}_{6}\;}{{}^{3}\!\!\diagup\!\!{}_{4}\;}=\frac{34}{9}=3.78m$

$\displaystyle {{l}_{2}}=\frac{0.5}{\tan {{\theta }_{2}}}=\frac{0.5}{\tan {{53}^{\circ }}}=\frac{0.5}{{}^{4}\!\!\diagup\!\!{}_{3}\;}=\frac{1.5}{4}=0.375m$

Thus the length of the light streak is, l₁ – l₂ = (3.78-0.375) = 3.405 m

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Laws of Reflection

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