+1-415-670-9521

info@expertsmind.com

# Refraction-Wave Theory Assignment Help

###
Optical Physics - Refraction-Wave Theory

**Refraction-Wave Theory**

AB is a plane wave front incident on **XY** at **∠BAA’ = ∠i. 1, 2, 3** are the corresponding incident rays normal to AB. According to Huygens principle, every point on AB is a source of secondary wavelets. Let the secondary wavelets from B strike XY at A’ in t seconds

Therefore** BA’ = c1 × t**

The secondary wavelets from A travel in the denser medium with a velocity **c**_{2} and would cover a distance**c**_{2} × t in t seconds. Therefore, with A as centre and radius equal to **c**_{2} × t, draw an arc B’.

From A’, draw a tangent plane touching the spherical arc tangentially at B’. therefore, A’B’ is the secondary wavefront after t sec. this would advance in the direction of rays, perpendicular to **A’B’**.

For A’B’ to be the true refracted wavefront, the secondary wavelets starting from any other point D on the incident wavefront AB, after refraction at P, must reach the point D’ on A’B’ in the same time in which the secondary wavelets from B reach A’. it is so, if time taken by secondary wavefront to travel a distance DP in rarer medium PD’ in denser medium = time taken by secondary wavelets to travel a distance **BA’** in rarer medium,

i.e. if **DP/c**_{1} + PD’/c_{2} = BA’/c_{1} (1)

Draw **PN ⊥ BA’ **

Therefore, **BA’ = BN + NA’ = DP + NA’**

Or **BA’/c**_{1} = DP/c_{1} + NA’/c_{1} (2)

From (1) and (2), we get,

**DP/c1 + NA’/c**_{1} = DP/c_{1 }+ PD’/c_{2}

Or, **NA’/c**_{1} = PD’/c_{2} (3)

Now, **Δs ABA’** and **PNA’** are similar,

Therefore, **AA’/PA’ = BA’/NA’ (4)**

Also, Δs AB’A’ and PD’A’ are similar,

Therefore, **AA’/PA’ = AB’/PD’ (5)**

From (4) and (5)

BA’/NA’ = AB’/PD’ or c_{1} × t/NA’ = c_{2} × t/PD’

Or, **NA’/c**_{1} = PD’/c_{2}, which is the necessary condition as per eqn. (3),

Therefore, **A’B’** is the true refracted wavefront. Let r be the angle of refraction. As angle of refraction is equal to the angle which the refracted plane wavefront A’B’ makes with the refracting surface **AA’**, therefore, **∠AA’B’ = r**.

Let **∠AA’B’ = r**, angle of refraction.

In **Δ AA’B’, sin i = BA’/AA’ = c**_{1} × t/AA’

In** Δ AA’B’, sin r = AB’/AA’ = c**_{2} × t/AA’

Therefore, **sin i/sin r = c**_{1}/c_{2} = µ

Hence, **µ = sin i/sin r**, which proves Snell’s law of refraction.

It is clear that the incident rays, normal to the interface XY and refracted ray, all lie in the same plane. (i.e. in the plane of the paper). This is the second law of refraction.

Hence laws of refraction are established on the basis of wave theory.

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

**Refraction-Wave Theory**

**XY**at

**∠BAA’ = ∠i. 1, 2, 3**are the corresponding incident rays normal to AB. According to Huygens principle, every point on AB is a source of secondary wavelets. Let the secondary wavelets from B strike XY at A’ in t seconds

Therefore

**BA’ = c1 × t**

The secondary wavelets from A travel in the denser medium with a velocity

**c**

_{2}and would cover a distance

**c**in t seconds. Therefore, with A as centre and radius equal to

_{2}× t**c**, draw an arc B’.

_{2}× tFrom A’, draw a tangent plane touching the spherical arc tangentially at B’. therefore, A’B’ is the secondary wavefront after t sec. this would advance in the direction of rays, perpendicular to

**A’B’**.

For A’B’ to be the true refracted wavefront, the secondary wavelets starting from any other point D on the incident wavefront AB, after refraction at P, must reach the point D’ on A’B’ in the same time in which the secondary wavelets from B reach A’. it is so, if time taken by secondary wavefront to travel a distance DP in rarer medium PD’ in denser medium = time taken by secondary wavelets to travel a distance

**BA’**in rarer medium,

i.e. if

**DP/c**

_{1}+ PD’/c_{2}= BA’/c_{1}(1)Draw

**PN ⊥ BA’**

Therefore,

**BA’ = BN + NA’ = DP + NA’**

Or

**BA’/c**

_{1}= DP/c_{1}+ NA’/c_{1}(2)From (1) and (2), we get,

**DP/c1 + NA’/c**

_{1}= DP/c_{1 }+ PD’/c_{2}

Or,

**NA’/c**

_{1}= PD’/c_{2}(3)Now,

**Δs ABA’**and

**PNA’**are similar,

Therefore,

**AA’/PA’ = BA’/NA’ (4)**

Also, Δs AB’A’ and PD’A’ are similar,

Therefore,

**AA’/PA’ = AB’/PD’ (5)**

From (4) and (5)

BA’/NA’ = AB’/PD’ or c

BA’/NA’ = AB’/PD’ or c

_{1}× t/NA’ = c_{2}× t/PD’Or,

**NA’/c**

_{1}= PD’/c_{2}, which is the necessary condition as per eqn. (3),

Therefore,

**A’B’**is the true refracted wavefront. Let r be the angle of refraction. As angle of refraction is equal to the angle which the refracted plane wavefront A’B’ makes with the refracting surface

**AA’**, therefore,

**∠AA’B’ = r**.

Let

**∠AA’B’ = r**, angle of refraction.

In

**Δ AA’B’, sin i = BA’/AA’ = c**

_{1}× t/AA’In

**Δ AA’B’, sin r = AB’/AA’ = c**

_{2}× t/AA’Therefore,

**sin i/sin r = c**

_{1}/c_{2}= µHence,

**µ = sin i/sin r**, which proves Snell’s law of refraction.

It is clear that the incident rays, normal to the interface XY and refracted ray, all lie in the same plane. (i.e. in the plane of the paper). This is the second law of refraction.

Hence laws of refraction are established on the basis of wave theory.

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