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# Reflection-Wave Theory Assignment Help

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Optical Physics - Reflection-Wave Theory

**Reflection-Wave Theory **

According to Huygens principle, every point on AB is a source of secondary wavelets. Let the secondary wavelets from B strike M1M2 at A’ in t seconds.

Therefore, **BA’ = c × t (1)**

Where c is the velocity of light in the medium, the secondary wavelets from A will travel the same distance **c × t **in the same time. Therefore, with A as the centre and** c × t **as radius, draw an arc B’ so that

**AB’ = c × t (2)**

From A’ draw a tangent plane A’B’ touching the spherical arc tangentially at B’. Therefore, A’B’ is the secondary wavefront after t seconds. This would advance in the direction of rays** 1’, 2’, 3’, **which are the corresponding reflected rays perpendicular to A’B’.

For A’B’ to be true reflected wavefront, secondary wavelets starting from any other point D on the incident wavefront AB, must reach the point D’ on A’B’, after reflection at P, and that too, in the same time as the secondary wavelets take to go from B to A’. For this,

**DP + PD’ = BA’ (1)**

To prove it, draw **PN ⊥ BA’**

Therefore, **DP = BN**

In **Δs ABA’ and AB’A’ = 90°**

And AA’ is common, therefore Δs are congruent.

Therefore, **∠BAA’ = ∠B’A’A (2)**

As **PN || AB**, therefore, **∠ BAA’ = ∠NPA’ (3)**

From (2) and (3),

**∠BAA’ = ∠B’A’A**

**= ∠D’A’P +∠ NPA’ (4)**

Now, in **Δs NPA’** and **D’PA’**

**∠NPA’ = ∠D’PA” (already proved)**

∠PNA’ = ∠PD’A’ = 90° and **PA’** is common.

Therefore, **Δs** are congruent.

Therefore, **NA’ = PD’**

As DP = BN, therefore,

**BN + NA’ = DP + PD’**

Or** BA’ = DP + PD’ **which is the necessary condition as per eqn. (1).

Hence, A’B’ is the true reflected wavefront. In fig.

Angle of incident,** i = ∠B’AA’**

And angle of reflection,** r = ∠B’A’A**

From (2), **∠B’AA’ = ∠B’A’A** i.e. **i = r**, which is the law of reflection.

Further, the incident wavefront AB, the reflecting surface **M1M**_{2} and the reflected wavefront A’B’ are the perpendicular to the plane of the paper. Therefore, incident ray, normal to the mirror **M**_{1}M_{2} and reflected ray all lie in the plane of the paper. This is the second law of reflection.

ExpertsMind.com - Physics Help, Optical Physics Assignments, Reflection-Wave Theory Assignment Help, Reflection-Wave Theory Homework Help, Reflection-Wave Theory Assignment Tutors, Reflection-Wave Theory Solutions, Reflection-Wave Theory Answers, Optical Physics Assignment Tutors

**Reflection-Wave Theory**

Therefore,

**BA’ = c × t (1)**

Where c is the velocity of light in the medium, the secondary wavelets from A will travel the same distance

**c × t**in the same time. Therefore, with A as the centre and

**c × t**as radius, draw an arc B’ so that

**AB’ = c × t (2)**

From A’ draw a tangent plane A’B’ touching the spherical arc tangentially at B’. Therefore, A’B’ is the secondary wavefront after t seconds. This would advance in the direction of rays

**1’, 2’, 3’,**which are the corresponding reflected rays perpendicular to A’B’.

For A’B’ to be true reflected wavefront, secondary wavelets starting from any other point D on the incident wavefront AB, must reach the point D’ on A’B’, after reflection at P, and that too, in the same time as the secondary wavelets take to go from B to A’. For this,

**DP + PD’ = BA’ (1)**

To prove it, draw

**PN ⊥ BA’**

Therefore,

**DP = BN**

In

**Δs ABA’ and AB’A’ = 90°**

And AA’ is common, therefore Δs are congruent.

Therefore,

**∠BAA’ = ∠B’A’A (2)**

As

**PN || AB**, therefore,

**∠ BAA’ = ∠NPA’ (3)**

From (2) and (3),

**∠BAA’ = ∠B’A’A**

**= ∠D’A’P +∠ NPA’ (4)**

Now, in

**Δs NPA’**and

**D’PA’**

**∠NPA’ = ∠D’PA” (already proved)**

**and**

∠PNA’ = ∠PD’A’ = 90°

∠PNA’ = ∠PD’A’ = 90°

**PA’**is common.

Therefore,

**Δs**are congruent.

Therefore,

**NA’ = PD’**

As DP = BN, therefore,

**BN + NA’ = DP + PD’**

Or

**BA’ = DP + PD’**which is the necessary condition as per eqn. (1).

Hence, A’B’ is the true reflected wavefront. In fig.

Angle of incident,

**i = ∠B’AA’**

And angle of reflection,

**r = ∠B’A’A**

From (2),

**∠B’AA’ = ∠B’A’A**i.e.

**i = r**, which is the law of reflection.

Further, the incident wavefront AB, the reflecting surface

**M1M**

_{2}and the reflected wavefront A’B’ are the perpendicular to the plane of the paper. Therefore, incident ray, normal to the mirror

**M**and reflected ray all lie in the plane of the paper. This is the second law of reflection.

_{1}M_{2}ExpertsMind.com - Physics Help, Optical Physics Assignments, Reflection-Wave Theory Assignment Help, Reflection-Wave Theory Homework Help, Reflection-Wave Theory Assignment Tutors, Reflection-Wave Theory Solutions, Reflection-Wave Theory Answers, Optical Physics Assignment Tutors