__Light Rays and Laws of Reflection:__

Have you ever wondered what makes you see yourself in a mirror? How do you see different objects around you? Well, the answer is pretty simple. All of these are possible due to reflection of light. But what exactly is meant by reflection of light? Let us try and figure it out.

Considering light as a ray, it always travels in a straight line. When it strikes the boundary separating two different media such as air and glass, a part of it is returned back into the same medium. This phenomenon is called ‘Reflection of light‘ which is governed by the law of reflection.

According to laws of reflection, angle of incidence is equal to angle of reflection. Furthermore, the ray of light striking the surface, the reflected ray and the normal drawn at the point of incidence of light lie in the same plane.

Let us now try to prove the laws of reflection. This would require a very fundamental principle of physics known as *Fermat’s principle* of least time.

According to *Fermat’s principle* of least time, light always takes the path which takes minimum time for it to reach from one point to another. As light is considered to travel in straight lines, so the minimum time taken by light is the one in which it has to cover the minimum distance.

To understand the path taken by light, let us imagine a scenario in which a bird is sitting on one pole and its nest is on another pole as shown. Now, there is a mouse running continuously between these two poles. If the bird flies only in straight lines then where should it catch the mouse in order to fly minimum distance to reach its nest?

The shortest path between any two points is the straight line that connects them. To solve it let us take the reflection of the pole Bas shown:

Now, considering \(\triangle BPO\) and \(\triangle B’PO\):

S. No. |
Reason |
Statement |

1. |
\(BP=B’P\) | B’P is reflection of BP |

2. |
\(∠BPO=~∠B’PO~=~90^{\circ}\) | Poles are at right angles w.r.t ground |

3. |
\(PO=PO\) | Common |

4. |
\(∆BPO≅∆B’PO\) | SAS axiom of congruency |

Considering \(∆B’PO\) and \(∆AOQ\)

S. No |
Reason |
Statement |

1. |
\(∠AOQ~=~∠POB’\) | Vertically opposite angles |

2. |
\(∠AQO~=~∠PB’O~=~90^{\circ}\) | Poles are at right angles w.r.t ground |

3. |
\(∆B’OP≈∆AOQ\) | AA axiom of similarity |

4. |
\(∆BOP≈∆AOQ\) | \(∆BPO≅∆B’PO\) |

The shortest distance between the points A and B’ is the straight line joining them which passes through point O. This means that the shortest distance between poles A and B will be the reflection of OB’ i.e. OB . Thus, the bird should catch the mouse at point O in order to cover the minimum distance.

In this scenario,the path opted by bird can be considered as a ray of light travelling from point to and passing through O to follow the quickest path. Using the above scenario let us try to prove the law of reflection.

If a normal is drawn at point O as shown, then

S. No |
Reason |
Statement |

1. |
\(∠PON~=~∠QON~=~90^{\circ}\) | Normal drawn at point O |

2. |
\(∠BOP~=~AOQ\) | \(∆BOP≈∆AOQ\) |

3. |
\(θ_i~=~θ_r\) | From statement 1 and 2 |

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It can be seen that the law of reflection holds true and \(θ_i~=~θ_r\).