 Physics

### Topic covered

color{blue}{star} REFRACTION AND REFLECTION OF PLANE WAVES USING HUYGENS PRINCIPLE
color{blue}{star} COHERENT AND INCOHERENT ADDITION OF WAVES

### REFRACTION AND REFLECTION OF PLANE WAVES USING HUYGENS PRINCIPLE

color{brown} {bbul{"Refraction of a plane wave"}}
color{blue} ✍️We will now use Huygens principle to derive the laws of refraction. Let PP′ represent the surface separating medium 1 and medium 2, as shown in Fig. 10.4. color{blue} ✍️Let v_1 and v_2 represent the speed of light in medium 1 and medium 2, respectively.
We assume a plane wavefront AB propagating in the direction A′A incident on the interface at an angle i as shown in the figure.
Let τ be the time taken by the wavefront to travel the distance BC. Thus

color{purple}{BC = v_1 τ}

color{blue} ✍️In order to determine the shape of the refracted wavefront, we draw a sphere of radius v_2τ from the point A in the second medium (the speed of the wave in the second medium is v_2).
Let CE represent a tangent plane drawn from the point C on to the sphere. Then, AE = v_2 τ and CE would represent the refracted wavefront. If we now consider the triangles ABC and AEC, we readily obtain

color {blue}{sin i = (BC)/(AC) = (V_1τ)/(AC)}

............(10.1)

color {blue}{sin r = (AE)/(AC) = (V_1τ)/(AC)}

...............(10.2) where i and r are the angles of incidence and refraction, respectively.

Thus we obtain

color {blue}{(sini)/(sinr) = (v_1)/(v_2)}

...........(10.3)

color{blue} ✍️From the above equation, we get the important result that if r < i (i.e., if the ray bends toward the normal), the speed of the light wave in the second medium (v_2) will be less then the speed of the light wave in the first medium (v_1).
This prediction is opposite to the prediction from the corpuscular model of light and as later experiments showed, the prediction of the wave theory is correct.

color{blue} ✍️Now, if c represents the speed of light in vacuum, then,

color {blue}{n_1 = C/(v_1)}

............(10.4)

and

color {blue}{n_2 = C/(v_2)}

............(10.5)

It is known as the refractive indices of medium 1 and medium 2, respectively. In terms of the refractive indices, Eq. (10.3) can be written as

color {blue}{n_1 sin i = n_2 sin r}

..............(10.6)

color{blue} ✍️This is the Snell’s law of refraction. Further, if λ1 and λ 2 denote the wavelengths of light in medium 1 and medium 2, respectively and if the distance BC is equal to λ 1 then the distance AE will be equal to λ 2 (because if the crest from B has reached C in time τ, then the crest from A should have also reached E in time τ ); thus,

color{purple}{(λ_1)/(λ_2) = (BC)/(AE) = (v_1)/(v_2)}
or

color {blue}{(v_1)/(λ_1) = (v_2)/(λ_2)}

............(10.7)

color{blue} ✍️The above equation implies that when a wave gets refracted into a denser medium (v_1 > v_2) the wavelength and the speed of propagation decrease but the frequency ν (= v/λ) remains the same.

color{brown} {bbul{"Refraction at a rarer medium"}} color{blue} ✍️We now consider refraction of a plane wave at a rarer medium, i.e., v_2 > v_1. Proceeding in an exactly similar manner we can construct a refracted wavefront as shown in Fig. 10.5. The angle of refraction will now be greater than angle of incidence; however, we will still have color{purple}{n_1 sin i = n_2 sin r} . We define an angle i_c by the following equation

color {blue}{sin i_c = (n_2)/(n_1)}

...........(10.8)

color{blue} ✍️Thus, if i = i_c then sin r = 1 and r = 90°. Obviously, for i > i_c, there can not be any refracted wave. The angle i_c is known as the critical angle and for all angles of incidence greater than the critical angle, we will not have any refracted wave and the wave will undergo what is known as total internal reflection.
The phenomenon of total internal reflection and its applications was discussed in Section 9.4.

color{brown} {bbul{"Reflection of a plane wave by a plane surface"}}
color{blue} ✍️We next consider a plane wave AB incident at an angle i on a reflecting surface MN. If v represents the speed of the wave in the medium and if τ represents the time taken by the wavefront to advance from the point B to C then the distance BC = v_τ

color{blue} ✍️In order the construct the reflected wavefront we draw a sphere of radius v_τ from the point A as shown in Fig. 10.6. Let CE represent the tangent plane drawn from the point C to this sphere. Obviously

color{purple}{A E = BC = vτ} color{blue} ✍️If we now consider the triangles EAC and BAC we will find that they are congruent and therefore, the angles i and r (as shown in Fig. 10.6) would be equal. This is the "law of reflection."

color{blue} ✍️Once we have the laws of reflection and refraction, the behaviour of prisms, lenses, and mirrors can be understood.

color{blue} ✍️Here we just describe the behaviour of the wavefronts as they undergo reflection or refraction. In Fig. 10.7(a) we consider a plane wave passing through a thin prism. color{blue} ✍️Clearly, since the speed of light waves is less in glass, the lower portion of the incoming wavefront (which travels through the greatest thickness of glass) will get delayed resulting in a tilt in the emerging wavefront as shown in the figure. In Fig. 10.7(b) we consider a plane wave incident on a thin convex lens; the central part of the incident plane wave traverses the thickest portion of the lens and is delayed the most.

color{blue} ✍️The emerging wavefront has a depression at the centre and therefore the wavefront becomes spherical and converges to the point F which is known as the focus. In Fig. 10.7(c) a plane wave is incident on a concave mirror and on reflection we have a spherical wave converging to the focal point F.

color{blue} ✍️In a similar manner, we can understand refraction and reflection by concave lenses and convex mirrors.

color{blue} ✍️From the above discussion it follows that the total time taken from a point on the object to the corresponding point on the image is the same measured along any ray. For example, when a convex lens focusses light to form a real image, although the ray going through the centre traverses a shorter path, but because of the slower speed in glass, the time taken is the same as for rays travelling near the edge of the lens.

### The doppler effect

color{blue} ✍️We should mention here that one should be careful in constructing the wavefronts if the source (or the observer) is moving. For example, if there is no medium and the source moves away from the observer, then later wavefronts have to travel a greater distance to reach the observer and hence take a longer time.

color{blue} ✍️The time taken between the arrival of two successive wavefronts is hence longer at the observer than it is at the source. Thus, when the source moves away from the observer the frequency as measured by the source will be smaller. This is known as the Doppler effect.

color{blue} ✍️Astronomers call the increase in wavelength due to doppler effect as red shift since a wavelength in the middle of the visible region of the spectrum moves towards the red end of the spectrum.
When waves are received from a source moving towards the observer, there is an apparent decrease in wavelength, this is referred to as blue shift.

color{blue} ✍️For velocities small compared to the speed of light, we can use the same formulae which we use for sound waves.
The fractional change in frequency Δν//ν is given by –v_("radial//c"), where v_("radial") is the component of the source velocity along the line joining the observer to the source relative to the observer; v_("radial") is considered positive when the source moves away from the observer. Thus, the Doppler shift can be expressed as:

color {blue}{(DeltaV)/V = - (V_(radial))/C}

..............(10.9)

color{blue} ✍️The formula given above is valid only when the speed of the source is small compared to that of light. A more accurate formula for the Doppler effect which is valid even when the speeds are close to that of light, requires the use of Einstein’s special theory of relativity.

color{blue} ✍️The Doppler effect for light is very important in astronomy. It is the basis for the measurements of the radial velocities of distant galaxies.
Q 3188378207 What speed should a galaxy move with respect to us so that the sodium line at 589.0 nm is observed at 589.6 nm?
Class 12 Chapter 10 Example 1 Solution:

Since νλ = (DeltaV)/V = - (Deltaλ)/λ (for small changes in ν and λ). For

Δλ = 589.6 – 589.0 = + 0.6 nm
we get [using Eq. (10.9)]

(DeltaV)/V = - (Deltaλ)/λ = - (v_(radial))/C

or, v_(radial) ≅+ C (0.6)/(589.0) = + 3.06 xx10^5 ms^(-1)

= 306 km//s

Therefore, the galaxy is moving away from us.
Q 3118378209 (a) When monochromatic light is incident on a surface separating two media, the reflected and refracted light both have the same frequency as the incident frequency. Explain why?
(b) When light travels from a rarer to a denser medium, the speed decreases. Does the reduction in speed imply a reduction in the energy carried by the light wave?
(c) In the wave picture of light, intensity of light is determined by the square of the amplitude of the wave. What determines the intensity of light in the photon picture of light.
Class 12 Chapter 10 Example 2 Solution:

(a) Reflection and refraction arise through interaction of incident light with the atomic constituents of matter. Atoms may be viewed as oscillators, which take up the frequency of the external agency (light) causing forced oscillations. The frequency of light emitted by a charged oscillator equals its frequency of oscillation. Thus, the frequency of scattered light equals the frequency of incident light.
(b) No. Energy carried by a wave depends on the amplitude of the wave, not on the speed of wave propagation.
(c) For a given frequency, intensity of light in the photon picture is determined by the number of photons crossing an unit area per unit time.

### COHERENT AND INCOHERENT ADDITION OF WAVES

color{blue} ✍️Here, we will discuss the interference pattern produced by the superposition of two waves.

color{blue} ✍️Indeed the entire field of interference is based on the superposition principle according to which at a particular point in the medium, the resultant displacement produced by a number of waves is the vector sum of the displacements produced by each of the waves. Consider two needles S_1 and S_2 moving periodically up and down in an identical fashion in a trough of water [Fig. 10.8(a)].

color{blue} ✍️They produce two water waves, and at a particular point, the phase difference between the displacements produced by each of the waves does not change with time; when this happens the two sources are said to be coherent. color{blue} ✍️Figure 10.8(b) shows the position of crests (solid circles) and troughs (dashed circles) at a given instant of time.

color{blue} ✍️Consider a point P for which S_1 P = S_2 P Since the distances S_1 P and S_2 P are equal, waves from S_1 and S_2 will take the same time to travel to the point P and waves that emanate from S_1 and S_2 in phase will also arrive, at the point P, in phase. Thus, if the displacement produced by the source S_1 at the point P is given by

color{purple}{y_1 = a cos ωt} then, the displacement produced by the source S_2 (at the point P) will also be given by

color{purple}{y_2 = a cos ωt}

color{blue} ✍️Thus, the resultant of displacement at P would be given by color{purple}{y = y_1 + y_2 = 2 a cos ωt}

color{blue} ✍️Since the intensity is the proportional to the square of the amplitude, the resultant intensity will be given by
color{purple}{I = 4 I_0}
color{blue} ✍️where I_0 represents the intensity produced by each one of the individual sources; I_0 is proportional to a_2.

color{blue} ✍️In fact at any point on the perpendicular bisector of S_1S_2, the intensity will be 4I_0. The two sources are said to interfere constructively and we have what is referred to as constructive interference.

color{blue} ✍️We next consider a point Q [Fig. 10.9(a)] for which color{purple}{S_2Q –S_1Q = 2λ}

color{blue} ✍️The waves emanating from S_1 will arrive exactly two cycles earlier than the waves from S_2 and will again be in phase [Fig. 10.9(a)]. Thus, if the displacement produced by S_1 is given by

color{purple}{y_1 = a cos ωt}

color{blue} ✍️then the displacement produced by S_2 will be given by
color{purple}{y_2 = a cos (ωt – 4π) = a cos ωt}

color{blue} ✍️where we have used the fact that a path difference of 2λ corresponds to a phase difference of 4π.

color{blue} ✍️The two displacements are once again in phase and the intensity will again be 4 I_0 giving rise to constructive interference. In the above analysis we have assumed that the distances S_1Q and S_2Q are much greater than d (which represents the distance between S_1 and S_2 so that although S_1Q and S_2Q are not equal, the amplitudes of the displacement produced by each wave are very nearly the same. We next consider a point R [Fig. 10.9(b)] for which S_2R – S_1R = –2.5λ

color{blue} ✍️The waves emanating from S_1 will arrive exactly two and a half cycles later than the waves from S_2 [Fig. 10.10(b)]. Thus if the displacement produced by S_1 is given by

color{purple}{y_1 = a cos ωt}

color{blue} ✍️then the displacement produced by S_2 will be given by
color{purple}{y_2 = a cos (ωt + 5π) = – a cos ωt}

color{blue} ✍️where we have used the fact that a path difference of 2.5λ corresponds to a phase difference of 5π.

color{blue} ✍️The two displacements are now out of phase and the two displacements will cancel out to give zero intensity. This is referred to as destructive interference. To summarise: If we have two coherent sources S_1 and S_2 vibrating in phase, then for an arbitrary point P whenever the path difference,

color {blue}{S_1P ~ S_2P = nλ \ \ \ \ (n = 0, 1, 2, 3,...)}

.............(10.10)

color{blue} ✍️we will have constructive interference and the resultant intensity will be 4I_0; the sign ~ between S_1P and S_2 Prepresents the difference between S_1P and S_2 P. On the other hand, if the point P is such that the path difference,

color {blue}{S_1P ~ S_2P = (n+1/2)λ \ \ \ \ (n = 0.1,2,3.)}

..........(10.11)

color{blue} ✍️we will have destructive interference and the resultant intensity will be zero. Now, for any other arbitrary point G (Fig. 10.10) let the phase difference between the two displacements be φ. Thus, if the displacement produced by S_1 is given by
y_1 = a cos ωt

color{blue} ✍️ then, the displacement produced by S_2 would be

color{purple}{y_2 = a cos (ωt + φ )}

color{blue} ✍️and the resultant displacement will be given by

color{purple}{y = y_1 + y_2}

color{purple}{= a [cos ωt + cos (ωt +phi )]}

color{purple}{= 2 a cos (φ//2) cos (ωt + φ//2)}

color{blue} ✍️The amplitude of the resultant displacement is 2a cos (φ//2) and therefore the intensity at that point will be

color {blue}{I = 4 I0 cos^2 (phi//2)}

...........(10.12)

color{blue} ✍️If φ = 0, ± 2 π, ± 4 π,… which corresponds to the condition given by Eq. (10.10) we will have constructive interference leading to maximum intensity. On the other hand, if φ = ± π, ± 3π, ± 5π … [which corresponds to the condition given by Eq. (10.11)]

color{blue} ✍️we will have destructive interference leading to zero intensity. Now if the two sources are coherent (i.e., if the two needles are going up and down regularly) then the phase difference φ at any point will not change with time and we will have a stable interference pattern; i.e., the positions of maxima and minima will not change with time.

color{blue} ✍️However, if the two needles do not maintain a constant phase difference, then the interference pattern will also change with time and, if the phase difference changes very rapidly with time, the positions of maxima and minima will also vary rapidly with time and we will see a “time-averaged” intensity distribution. When this happens, we will observe an average intensity that will be given by

color {blue}{ I = 4I_0 cos^2 φ//2)}

..................(10.13)

color{blue} ✍️ where angular brackets represent time averaging. Indeed it is shown in Section 7.2 that if φ(t) varies randomly with time, the time-averaged quantity cos^2 (φ//2)  will be 1//2.

color{blue} ✍️This is also intuitively obvious because the function cos^2 (φ//2) will randomly vary between 0 and 1 and the average value will be 1//2. The resultant intensity will be given by

color {blue}{I= 2 I_0}

..............(10.14) at all points.

color{blue} ✍️When the phase difference between the two vibrating sources changes rapidly with time, we say that the two sources are incoherent and when this happens the intensities just add up. This is indeed what happens when two separate light sources illuminate a wall. 