The Angle of incidence can be defined as the angle with which a light ray strikes a reflecting surface. But the angle of strike is measured as the angle between the ray and the normal to the surface at the point of incidence of the ray. It has a relation with the angle of reflection and also with the type of reflecting surface.

Related Calculators | |

Angle Calculator | Side Angle side Calculator |

Angle between Two Vectors Calculator | Complementary Angle Calculator |

Look at the below diagram.

Look at the above diagram. Let a right ray AO from a source A strike a plane mirror at O and reflected as ray OB. Let B be a point on ray OB and let ON be the normal to the mirror at O. The angle of incidence is ‘x’ and angle of reflection is ‘y’, as shown. Let ‘t’ be the time taken for the ray to travel the distance AO + OB and ‘c’ be the velocity of light, which is a constant.

The time ‘t’ can be worked out as,

**$The\ Angle\ of\ Incidence$ = $Angle\ of\ Reflection$**

Take a white card board sheet and fix the same firmly on a board. Draw a thick line MM’ and mark the midpoint O. Draw a perpendicular ON at O. Place a plain mirror on MM’ with its center at O.Take three pins and fix them vertically on the sheet at positions C, B and A on one side of ON such that the pins lie on a straight line when viewed from A. Now move to the other side of the perpendicular ON and fix three more pins at F, E and D in such way that the images A’ B’ C’ of the pins at A, B, C and the pins at D, E, F are all in one straight line. This is the most important point of the entire experiment.

Now remove the mirror and the pins carefully marking the points where they have been fixed. At this stage we will not be able to locate the positions of A’, B’ and C’. Draw a perpendicular AP from A and extend that line. Join DO and extend that too. Let A’ be the point of intersection of these extended line. A’ is nothing but the position of the image that appeared when the mirror was there. Same way locate the positions of B’ and C’.

If you join AO and measure the angles AON and DON, you will find them to be equal. Thus, it proves the angle of incidence equals angle of reflection. Also you will notice that the following measures to be equal. AP = PA’, BQ = QB’ and CR = RC’, which proves the image distance equals the object distance.

The ray I from a light source strikes the mirror MM’ at point O This ray is called as incident ray. ON is the perpendicular to the mirror MM’ at O which is also called the normal to the mirror at O. The angle ‘x’ made by the ray I, with the normal ON is called the ‘angle of incidence’.

The incident ray, is reflected by the mirror when it strikes the mirror at O and the ray after reflection ‘R’ is called as ‘**reflected ray**’. The angle ’y’ made by the reflected ray with the same ray is called ‘angle of reflection’.

As per law of reflection the angle of incidence is always equal to angle of reflection. This can be proved mathematically and also by a simple experiment. There are many mathematical proofs but the simplest proof is based on **Fermat’s Principle**.

The time ‘t’ can be worked out as,

$t$ = $\frac{AO}{c}$ + $\frac{OB}{c}$ or,

$t$ = $\frac{\sqrt{a^{2}+p^{2}}}{c}+\frac{\sqrt{(l-a)^{2}+q^{2}}}{c}$

$t$ = $\frac{\sqrt{a^{2}+p^{2}}}{c}+\frac{\sqrt{(l-a)^{2}+q^{2}}}{c}$

As per Fermat's principle light takes the path of least time. Therefore, the derivative $\frac{dt}{da}$ must be 0.

That is, $0$ = $\frac{1}{c}[\frac{a}{\sqrt{a^{2}+p^{2}}}+\frac{-(l-a)}{\sqrt{(l-a)^{2}+q^{2}}}]$

or, $\frac{a}{\sqrt{a^{2}+p^{2}}}$ = $\frac{(l - a)}{\sqrt{(l - a)^{2} + q^{2}}}$

But as per trigonometry, this is same as $\sin (x)$ = $\sin(y)$ or $x$ = $y$.

Hence,** **

The law of reflection can also be proved experimentally. The arrangement for the experiment is described below.

Take a white card board sheet and fix the same firmly on a board. Draw a thick line MM’ and mark the midpoint O. Draw a perpendicular ON at O. Place a plain mirror on MM’ with its center at O.Take three pins and fix them vertically on the sheet at positions C, B and A on one side of ON such that the pins lie on a straight line when viewed from A. Now move to the other side of the perpendicular ON and fix three more pins at F, E and D in such way that the images A’ B’ C’ of the pins at A, B, C and the pins at D, E, F are all in one straight line. This is the most important point of the entire experiment.

Now remove the mirror and the pins carefully marking the points where they have been fixed. At this stage we will not be able to locate the positions of A’, B’ and C’. Draw a perpendicular AP from A and extend that line. Join DO and extend that too. Let A’ be the point of intersection of these extended line. A’ is nothing but the position of the image that appeared when the mirror was there. Same way locate the positions of B’ and C’.

If you join AO and measure the angles AON and DON, you will find them to be equal. Thus, it proves the angle of incidence equals angle of reflection. Also you will notice that the following measures to be equal. AP = PA’, BQ = QB’ and CR = RC’, which proves the image distance equals the object distance.

Related Topics | |

Physics Help | Physics Tutor |