Explain the refraction of light on the basis of wave theory and also prove the laws of refraction of light.

i) The incident rays, refracted rays and normal lie in the equivalent plane.

ii) The incident ray and refracted ray lie on opposite sides of normal. 

iii) The ratio of velocity of light in rarer medium to velocity of light in denser medium is a constant called refractive index of denser medium with respect to rarer medium.

Explanation laws of refraction :

i) Let consider XY be the plane refracting surface separating two media air and glass of refractive indices μ1 and μ2 serially.

ii) A plane wavefront AB is developing at an angle towards XY from the air medium. It is bounded by the rays AA1 and BB1 which are incident rays.

iii) When point 'A' reaches 'A1' then point 'B' will be at 'P'. It even now has to cover distance PB1 to reach XY.

iv) According to Huygens' principle the secondary waves will originate from A1 and will spread over a hemisphere in the glass.

v) All the rays between AA1 and BB1 will reach XY and spread over the hemispheres of increasing radii in the glass. The surface of tangential of all such hemispheres is RB1. This allow to rise refracted wavefront B1R in the glass.

vi)  A1R and B1R1, are refracted rays.

vii) Let consider c1 and c2 be the velocities of light in air and glass .

viii) At any instant of time 't' distance covered by the incident wavefront from point P to B1 = PB1 = c1t.

The distance covered by the secondary wave from point A1 to R = A1R = c2t. 

Proof of laws of refraction:

i)  From figure,

angle AA1M + angle MA1P = 90° .................................(1)

and angle MA1P+ angle PA1B1 = 90° ...........................(2)

From equations (1) and (2),

angle AA1M = angle PA1B1 = i 

ii) Similarly, angle NA1R = angle N1B1R1 = r

We have, angle N1B1R1 + angle N1B1R = 90°.............(3)

 and angle N1B1R + angle A1B1R = 90° ......................(4)

From equations (3) and (4)

angle N1B1R1 = angle A1B1R = r

iii) In triangle A 1PB1, sin i = PB1/A1B1 = c1t/A1B1................(5)

iv) In triangle A1RB1, sin r = A1R/A1B1 = c2t /A1B1..............(6)

v) Dividing equation (5) by (6),

Therefore sin i/ sin r = c1/ c2....................(7)

Also c1/ c2 = μ2/ μ1 = 1μ2 ......................(8)

Where as 1μ2 = R.I. of water with respect to air,

sin i/ sin r = μ2/ μ1

vi) From the explanation, it is clear that incident rays AA, BB, refracted rays A,R, B,R and normal MN and MIN lie on the same plane XY. Also incident ray AA, and refracted ray AR lie on opposite sides of normal MN. Therefore, laws of refraction can be explained.

THE END

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With the help of a neat diagram | explain the reflection of light from the plane reflecting surface on the basis of wave theory of light.

So here first of all we know that, 

What is the laws of reflection :

i) In the laws of reflection, the incident rays, reflected rays, and normal to the reflecting surface at the point of incidence, all lie in the same level.

ii) In the laws of reflection, the incident rays and the reflected rays lie on the opposite sides of the normal.

iii) The angle of incidence is equal or we can say equivalent to the angle of reflection.

For clear understanding the law of reflection in simple and easy way. So for that lets see the simple explanation about all these points.

Explanation :

i) The plane wavefront AB is developing at an angle towards plane reflecting surface XY.AA1 and BB1 are incident rays.

ii) When point 'A' come to XY at A1, than ray at 'B' reaches at point P and it has to cover distance PB1 to reach the reflecting surface XY.

iii) Let's consider 't' be the time to  required to cover distance PB1. 

In this time meantime, the secondary waves are emitted from the point A1 and it will spread over the hemisphere of the radius A1R in the same medium.

The distance covered by the secondary waves to reach from A1 to R in time 't' is same as the distance covered by the primary waves to reach from point P to B1.

So therefore A1R = PB1 = ct.

iv) All other rays in between AA1 and BB1 will reach XY after A1 and before B1. Therefore they will also emits secondary waves of the decreasing radii.

v) The surface touch all such hemispheres is RB1 which is reflected wavefront, bounded by the reflected rays A1R and B1Q.

vi) Let's draw A1M perpendicular to XY and B1N perpendicular to XY.

Thus,  the angle of incidence is angle AA1M = angle BB1N = i and the angle of reflection is angle MA1R = angle NB1Q = r.

angle RA1B1 = 90 - r

angle PB1A1 = 90 - i

vii) In the triangle A1RB1 and triangle A1PB1

Here angle A1RB1 congruent to angle A1PB1

A1R = PB1,............ (Reflected waves travel in equal in same medium in equal time).

A1B1 = A1B1.........(Common side) 

Therefore triangle A1RB1 congruent to triangle A1PB1

Therefore angle RA1B1 = angle PB1A1

Therefore  90 - r = 90 - i

Therefore  i = r

viii) From the above given figure, it clear that the incident ray, reflected ray and normal lie in the same level or plane.

ix) This laws of reflection of light from the plane reflecting surface on the basis of Huygen's wave theory.

THE END

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The Huygen's construction of spherical wavefront in simple and easy way !!!

Here we come to know about the Huygen's construction of spherical wavefront.

So let's come to know,

i) The spherical wavefront is formed when the source of light is at the finite distance from the point of observation.

ii) So let's take 'S' be the point source of light in the air.

The point PQR represents spherical wavefront at any instant. 

The wavefront PQR is act as a primary wave, which is propagated through the air.

iii) So according to the Huygens principle, the all points on the PQR will act as a secondary source of light and sending out secondary wavelets with the same velocity 'c' in the air.

iv) To find out new wavefront at a later instant 't',

So for our clear understanding lets draw the hemisphere with the points P, Q, R....as centres and 'ct' as radius in the forward direction.

v) The surface tangential to all such hemisphere is an envelope at that instant 't'.

Such a surface is passing through the points P1Q1R1....so on the hemispheres and touching to all the hemispheres.

This surface is the new wavefront at that instant 't'.

vi) SPN1, SQN2, SRN3 these are the wave normal at points P, Q, R respectively.

vii) These wave normal show the direction of propagation of the spherical wavefront.

viii) The new wavefront P1Q1R1 is parallel to the points P, Q, R at every instant.

THE END

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 Explain the Huygen's construction of plane wavefront in easy and simple way !!!

Here we know all about, the construction of the plane wavefront.

So let's come to know,

i) The plane wavefront is formed when the point of observation is very far away or we can say at infinity from the primary source of light.

ii) So here let's PQR represent a plane wavefront at any instant.

So according to the Huygen's principle, all the points on this wavefront will act as a secondary source of light and sending out secondary wavelets in the forward direction only.

iii) For make it to undestand easily, first of all we need to draw hemisphere with Point P,Q,R.... as centres and 'ct' as a radius.

The surface tangential to all such hemisphere is P1Q1R1....at instant 't'.

The new wavefront at time 't'.

iv) The plane wavefront is propagated or travel as a plane waves in the homogeneous isotropic medium. Therefore they are parallel to each other.

v) PP1N1, QQ1N2, RR1N3 these are the wave normal at point P,Q,R respectively.

These wave normal shows the direction of propagation of the plane wavefront.

vi) The new wavefront P1Q1R1 is parallel to the primary wavefront PQR.

THE END

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Difference in between primary source of light and secondary source of light in easy way !!!

Here we know about the distinguish/ difference between primary source of light and the secondary source of light.

Primary source of light

i) The primary source of light is a real source of light.

ii) Primary source of light sends out primary waves in the all possible directions.

iii) The primary source of light is effective at every point on it's surface.

iv) The primary source of light is situated in the air.

The secondary source of light 

i) The secondary source of light is a fictitious source of light.

ii) The secondary source of light sends out secondary waves only in the forward direction.

iii) The secondary waves is effective only at the points where it touches the envelope.

iv) The secondary source of light is situated on the wavefront.

THE END

State what are the Huygen's principle statement !!!

Hello dear all readers,

Welcome to this concept, so here in this concept you come to know about the principle statement of Huygen's. 

So lets start to know,

i) Every point on the primary wavefront it acts as a secondary waves or wavelets in the all possible directions.

ii) The new generated secondary wave is quite more effectiveness in the forward direction only.

iii) The resultant wavefront at any position is given by the tangent to all the secondary waves or wavelets at that instant.

So we hope you understand this concept.

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State and explain the main characteristics of the wavefront in easy and simply way.

Hello dear all readers,

Welcome to this new concept, so here in this concept you come to know about what are the main characteristics of the wavefront with explanation in easy and simply way.

So let's start to our topic !!!

i) The wavefronts travel or propagate as much as speed of light in all the direction in the isotropic medium.

Explanation : Here in this statement, it is clearly tell that the wavefront is able to travel or propagate with speed of light in the isotropic medium.

As you know that, the speed of light is 3×10^8  metre 

per second.

ii) The phase difference between any two points in the same consecutive wavefronts is 2π.

So that, if the phase at the crest is 2π, than the phase at the next consecutive crest is 4π and like wise so on.

So therefore, at any crest the phase is (2nπ) and the phase at any trough is (2n+1).

Where 'n' represent as integer.

iii) The wavefronts always travel or propagate in the forward direction only.

iv) In during the propagation of the spherical wavefront from the source than slowly slowly wave become weaker.

Reason for that is : The same energy is distributed over whole circumference of the circle.

So as the circle of radii get increasing than the wavefronts become slowly slowly weaker.

Let's take an one example on it : When you through stone in the pond or in the lake than, when the stone goes in the water, at that place ripples is get generated isn't it ?

After ripples get generated at specific point than slowly slowly it become less energetic, isn't it ?

So like wise here also in the spherical wavefront.

v) So let's know what is the response of wavefront in anisotropic medium. In the anisotropic medium, the wavefront travels or propagate with different different velocities in different different directions, it is because of variation in the densities of the medium.

So we hope you understand this concept.

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What is the expecting wavefront shape when the portion of the wavefront of light from a distant star intercepted by the earth ?

Hello dear all readers,

Welcome to this new concept, so here in this concept you come to know about 'what is the expecting wavefront shape when the portion of the wavefront of light from a distant star intercepted by the earth ?

So let's come to our topic !!!

When the portion of the wavefront of light from the distant or faraway star intercepted by the earth ? than what should be it's geometrical wavefront shape ?

Here you know all about this !!!

As we already know that the star has it's own light means that it is one type of source of light which is very far, that is infinity.

So that when the light coming from the star, which is intercepted by the earth is a plane wavefront.

As you can see in the above given image or figure.

So we hope you understand this concept very clearly and nicely with great interest mode.

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What is the expecting shape of the wavefront when light emerging out of a convex lens ?

Hello dear all readers,

Welcome to this new concept, so here in this concept you come to know 'about what is the shape of wavefront when light emerging out of a convex lens'.

So let's come to our topic !!!

When light emerging out of a convex lens than the shape of the wavefront is plane.

Explanation : When a point source placed at the focus of a convex lens, than whatever rays emerging or come out from the lens is in parallel form.

So that, the wavefront is a plane wavefront.

Here in the given above image you can see very clearly, how these rays emerging in a parallel form in a plane wavefront.

So we hope you understand this concept very clearly and simply.

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