Mathematics SEEING THE SINGLE SLIT DIFFRACTION PATTERN, POLARISATION FOR CBSE NCERT -4

### Topic covered

color{blue}{star} SEEING THE SINGLE SLIT DIFFRACTION PATTERN
color{blue}{star} POLARISATION

### SEEING THE SINGLE SLIT DIFFRACTION PATTERN

color{blue} ✍️It is surprisingly easy to see the single-slit diffraction pattern for oneself. The equipment needed can be found in most homes two razor blades and one clear glass electric bulb preferably with a straight filament.

color{blue} ✍️One has to hold the two blades so that the edges are parallel and have a narrow slit in between. This is easily done with the thumb and forefingers (Fig. 10.18).

color{blue} ✍️Keep the slit parallel to the filament, right in front of the eye. Use spectacles if you normally do. With slight adjustment of the width of the slit and the parallelism of the edges, the pattern should be seen with its bright and dark bands.

color{blue} ✍️Since the position of all the bands (except the central one) depends on wavelength, they will show some colours. Using a filter for red or blue will make the fringes clearer. With both filters available, the wider fringes for red compared to blue can be seen.

color{blue} ✍️ In this experiment, the filament plays the role of the first slit S in Fig. 10.16. The lens of the eye focuses the pattern on the screen (the retina of the eye). With some effort, one can cut a double slit in an aluminium foil with a blade.

color{blue} ✍️The bulb filament can be viewed as before to repeat Young’s experiment. In daytime, there is another suitable bright source subtending a small angle at the eye. This is the reflection of the Sun in any shiny convex surface (e.g., a cycle bell). Do not try direct sunlight – it can damage the eye and will not give fringes anyway as the Sun subtends an angle of (1/2)º.

color{blue} ✍️In interference and diffraction, light energy is redistributed. If it reduces in one region, producing a dark fringe, it increases in another region, producing a bright fringe. There is no gain or loss of energy, which is consistent with the principle of conservation of energy.

color{brown} {bbul{"Resolving power of optical instruments"}}
color{blue} ✍️As we had discussed about telescopes. The angular resolution of the telescope is determined by the objective of the telescope. The stars which are not resolved in the image produced by the objective cannot be resolved by any further magnification produced by the eyepiece.

color{blue} ✍️The primary purpose of the eyepiece is to provide magnification of the image produced by the objective. Consider a parallel beam of light falling on a convex lens. If the lens is well corrected for aberrations, then geometrical optics tells us that the beam will get focused to a point.

color{blue} ✍️However, because of diffraction, the beam instead of getting focused to a point gets focused to a spot of finite area. In this case the effects due to diffraction can be taken into account by considering a plane wave incident on a circular aperture followed by a convex lens (Fig. 10.19).

color{blue} ✍️The analysis of the corresponding diffraction pattern is quite involved; however, in principle, it is similar to the analysis carried out to obtain the single-slit diffraction pattern. Taking into account the effects due to diffraction, the pattern on the focal plane would consist of a central bright region surrounded by concentric dark and bright rings (Fig. 10.19).

color{blue} ✍️A detailed analysis shows that the radius of the central bright region is approximately given by

color {blue}{r_0 approx (1.22λf)/(2a) = (0.6λf)/a}

............(10.24)

color{blue} ✍️where f is the focal length of the lens and 2a is the diameter of the circular aperture or the diameter of the lens, whichever is smaller. Typically if

color{purple}{λ ≈ 0.5 μm, f ≈ 20 cm" and "a ≈ 5 cm}

color{blue} ✍️we have r_0 ≈ 1.2 μm

color{blue} ✍️Although the size of the spot is very small, it plays an important role in determining the limit of resolution of optical instruments like a telescope or a microscope. For the two stars to be just resolved

color{purple}{f Delta theta approx r_0 = (0.61λf)/a}

implying

color {blue}{Delta theta approx = (0.61λf)/a}

...............(10.25)

color{blue} ✍️Thus Δθ will be small if the diameter of the objective is large. This implies that the telescope will have better resolving power if a is large. It is for this reason that for better resolution, a telescope must have a large diameter objective.

color{blue} ✍️We can apply a similar argument to the objective lens of a microscope. In this case, the object is placed slightly beyond f, so that a real image is formed at a distance v [Fig. 10.20]. The magnification – ratio of image size to object size – is given by m l v//f. It can be seen from Fig. 10.20 that

color {blue}{D/f = 2 tan β}

..............(10.26)

color{blue} ✍️where 2β is the angle subtended by the diameter of the objective lens at the focus of the microscope.

color{blue} ✍️When the separation between two points in a microscopic specimen is comparable to the wavelength λ of the light, the diffraction effects
become important. The image of a point object will again be a diffraction pattern whose size in the image plane will be

color {blue}{v theta=v ((1.22λ)/D)}

................(10.27)

color{blue} ✍️Two objects whose images are closer than this distance will not be resolved, they will be seen as one. The corresponding minimum separation, d_(min,) in the object plane is given by

color{purple}{d_(min) = [ v((1.22λ)/D)]/m}

color{purple}{=(1.22λ)/D.v/m}

color {blue}{=(1.22fλ)/D}

....................(10.28)

color{blue} ✍️Now, combining Eqs. (10.26) and (10.28), we get

color{purple}{d_(min) = (1.22fλ)/(2 tan beta)}

color {blue}{= (1.22λ)/(2 sin beta)}

.............(10.29)

color{blue} ✍️If the medium between the object and the objective lens is not air but a medium of refractive index n, Eq. (10.29) gets modified to

color {blue}{d_(min) = (1.22fλ)/(2 sin beta)}

.............(10.30)

color{blue} ✍️The product nsinβ is called the numerical aperture and is sometimes marked on the objective.

color{blue} ✍️The resolving power of the microscope is given by the reciprocal of the minimum separation of two points seen as distinct. It can be seen from Eq. (10.30) that the resolving power can be increased by choosing a medium of higher refractive index.

color{blue} ✍️Usually an oil having a refractive index close to that of the objective glass is used. Such an arrangement is called an ‘oil immersion objective’. Notice that it is not possible to make sinβ larger than unity.

color{blue} ✍️Thus, we see that the resolving power of a microscope is basically determined by the wavelength of the light used. There is a likelihood of confusion between resolution and magnification, and similarly between the role of a telescope and a microscope to deal with these parameters.

color{blue} ✍️A telescope produces images of far objects nearer to our eye. Therefore objects which are not resolved at far distance, can be resolved by looking at them through a telescope.

color{blue} ✍️A microscope, on the other hand, magnifies objects (which are near to us) and produces their larger image. We may be looking at two stars or two satellites of a far-away planet, or we may be looking at different regions of a living cell. In this context, it is good to remember that a telescope resolves whereas a microscope magnifies.

color{brown} {bbul{"The validity of ray optics"}}
color{blue} ✍️An aperture (i.e., slit or hole) of size a illuminated by a parallel beam sends diffracted light into an angle of approximately ≈ λ/a.

color{blue} ✍️This is the angular size of the bright central maximum. In travelling a distance z, the diffracted beam therefore acquires a width zλ/a due to diffraction.

color{blue} ✍️It is interesting to ask at what value of z the spreading due to diffraction becomes comparable to the size a of the aperture. We thus approximately equate zλ/a with a. This gives the distance beyond which divergence of the beam of width a becomes significant. Therefore,

color {blue}{z = (a^2)/λ}

.............(10.31)

color{blue} ✍️We define a quantity z_F called the Fresnel distance by the following equation

color{purple}{Z_F = a^2 //λ}

color{blue} ✍️Equation (10.31) shows that for distances much smaller than z_F, the spreading due to diffraction is smaller compared to the size of the beam.

color{blue} ✍️It becomes comparable when the distance is approximately z_F. For distances much greater than z_F, the spreading due to diffraction dominates over that due to ray optics (i.e., the size a of the aperture). Equation (10.31) also shows that ray optics is valid in the limit of wavelength tending to zero.
Q 3128078801

Assume that light of wavelength 6000Å is coming from a star. What is the limit of resolution of a telescope whose objective has a diameter of 100 inch?
Class 12 Chapter 10 Example 6
Solution:

A 100 inch telescope implies that 2a = 100 inch = 254 cm. Thus if,

λ ≈ 6000Å = 6×10^(–5) cm

then

Deltathetaapprox (0.61xx6xx10^(-5))/(127) ≈ 2.9 xx 10^(-7) radians
Q 3168078805

For what distance is ray optics a good approximation when the aperture is 3 mm wide and the wavelength is 500 nm?
Class 12 Chapter 10 Example 7
Solution:

Z_F = (a^2)/λ = ((3xx10^(-3))^2)/(5xx10^(-7) = 18m

This example shows that even with a small aperture, diffraction spreading can be neglected for rays many metres in length. Thus, ray optics is valid in many common situations.

### POLARISATION

color{blue} ✍️Consider holding a long string that is held horizontally, the other end of which is assumed to be fixed.
If we move the end of the string up and down in a periodic manner, we will generate a wave propagating in the +x direction (Fig. 10.21). Such a wave could be described by the following equation

color {blue}{y (x,t ) = a sin (kx – ωt)}

..............(10.32)

color{blue} ✍️where a and ω (= 2πν ) represent the amplitude and the angular frequency of the wave, respectively; further,

color {blue}{λ= (2pi)k}

............(10.33)

color{blue} ✍️represents the wavelength associated with the wave.

color{blue} ✍️Since the displacement (which is along the y direction) is at right angles to the direction of propagation of the wave, we have what is known as a transverse wave.

color{blue} ✍️Also, since the displacement is in the y direction, it is often referred to as a y-polarised wave. Since each point on the string moves on a straight line, the wave is also referred to as a linearly polarised wave.

color{blue} ✍️Further, the string always remains confined to the x-y plane and therefore it is also referred to as a plane polarised wave. In a similar manner we can consider the vibration of the string in the x-z plane generating a z-polarised wave whose displacement will be given by

color {blue}{z (x,t ) = a sin (kx – ωt )}

.................(10.34)

color{blue} ✍️It should be mentioned that the linearly polarised waves [described by Eqs. (10.33) and (10.34)] are all transverse waves; i.e., the displacement of each point of the string is always at right angles to the direction of propagation of the wave.

color{blue} ✍️Finally, if the plane of vibration of the string is changed randomly in very short intervals of time, then we have what is known as an unpolarised wave.

color{blue} ✍️Thus, for an unpolarised wave the displacement will be randomly changing with time though it will always be perpendicular to the direction of propagation.

color{blue} ✍️Light waves are transverse in nature; i.e., the electric field associated with a propagating light wave is always at right angles to the direction of propagation of the wave.

color{blue} ✍️This can be easily demonstrated using a simple polaroid. You must have seen thin plastic like sheets, which are called polaroids. A polaroid consists of long chain molecules aligned in a particular direction.

color{blue} ✍️The electric vectors (associated with the propagating light wave) along the direction of the aligned molecules get absorbed. Thus, if an unpolarised light wave is incident on such a polaroid then the light wave will get linearly polarised with the electric vector oscillating along a direction perpendicular to the aligned molecules; this direction is known as the pass-axis of the polaroid.

color{blue} ✍️Thus, if the light from an ordinary source (like a sodium lamp) passes through a polaroid sheet P_1, it is observed that its intensity is reduced by half. Rotating P_1 has no effect on the transmitted beam and transmitted intensity remains constant.

color{blue} ✍️Now, let an identical piece of polaroid P_2 be placed before P_1. As expected, the light from the lamp is reduced in intensity on passing through P_2 alone. But now rotating P_1 has a dramatic effect on the light coming from P_2. In one position, the intensity transmitted by P_2 followed by P_1 is nearly zero. When turned by 90º from this position, P_1 transmits nearly the full intensity emerging from P_2 (Fig. 10.22).

color{blue} ✍️The above experiment can be easily understood by assuming that light passing through the polaroid p_2 gets polarised along the pass-axis of p_2. If the pass-axis of p_2 makes an angle θ with the pass-axis of p_1, then when the polarised beam passes through the polaroid p_2, the component E cos θ (along the pass-axis of p_2) will pass through p_2. Thus, as we rotate the polaroid p_1 (or p_2), the intensity will vary as:

color {blue}{I = I_0 cos2θ}

..........(10.35)

color{blue} ✍️where I_0 is the intensity of the polarized light after passing through p_1.

color{blue} ✍️This is known as Malus’ law. The above discussion shows that the intensity coming out of a single polaroid is half of the incident intensity. By putting a second polaroid, the intensity can be further controlled from 50% to zero of the incident intensity by adjusting the angle between the pass-axes of two polaroids.

color{blue} ✍️Polaroids can be used to control the intensity, in sunglasses, windowpanes, etc. Polaroids are also used in photographic cameras and 3D movie cameras.

color{brown} {bbul{"Polarisation by scattering"}}
color{blue} ✍️The light from a clear blue portion of the sky shows a rise and fall of intensity when viewed through a polaroid which is rotated. This is nothing but sunlight, which has changed its direction (having been scattered) on encountering the molecules of the earth’s atmosphere.

color{blue} ✍️As Fig. 10.23(a) shows, the incident sunlight is unpolarised. The dots stand for polarisation perpendicular to the plane of the figure. The double arrows show polarisation in the plane of the figure. (There is no phase relation between these two in unpolarised light).

color{blue} ✍️Under the influence of the electric field of the incident wave the electrons in the molecules acquire components of motion in both these directions. We have drawn an observer looking at 90° to the direction of the sun.

color{blue} ✍️Clearly, charges accelerating parallel to the double arrows do not radiate energy towards this observer since their acceleration has no transverse component. The radiation scattered by the molecule is therefore represented by dots. It is polarised perpendicular to the plane of the figure. This explains the polarisation of scattered light from the sky.

color{blue} ✍️The scattering of light by molecules was intensively investigated by C.V. Raman and his collaborators in Kolkata in the 1920s. Raman was awarded the Nobel Prize for Physics in 1930 for this work.

color{brown} {bbul{"Polarisation by reflection"}}
color{blue} ✍️Figure 10.23(b) shows light reflected from a transparent medium, say, water. As before, the dots and arrows indicate that both polarisations are present in the incident and refracted waves.

color{blue} ✍️We have drawn a situation in which the reflected wave travels at right angles to the refracted wave. The oscillating electrons in the water produce the reflected wave.

color{blue} ✍️These move in the two directions transverse to the radiation from wave in the medium, i.e., the refracted wave. The arrows are parallel to the direction of the reflected wave. Motion in this direction does not contribute to the reflected wave.

color{blue} ✍️As the figure shows, the reflected light is therefore linearly polarised perpendicular to the plane of the figure (represented by dots). This can be checked by looking at the reflected light through an analyser. The transmitted intensity will be zero when the axis of the analyser is in the plane of the figure, i.e., the plane of incidence.

color{blue} ✍️When unpolarised light is incident on the boundary between two transparent media, the reflected light is polarised with its electric vector perpendicular to the plane of incidence when the refracted and reflected rays make a right angle with each other.

color{blue} ✍️Thus we have seen that when reflected wave is perpendicular to the refracted wave, the reflected wave is a totally polarised wave. The angle of incidence in this case is called Brewster’s angle and is denoted by i_B. We can see that i_B is related to the refractive index of the denser medium.

color{blue} ✍️Since we have iB+r = π//2, we get from Snell’s law

color{purple}{mu = (sini_B)/(sin r) = (sini_B) /(sin (pi//2-i_B))}

color {blue}{(sini_B)/(sin_B) = tani_B}

...........(10.36) This is known as "Brewster’s law."

color{blue} ✍️For simplicity, we have discussed scattering of light by 90º, and reflection at the Brewster angle. In this special situation, one of the two perpendicular components of the electric field is zero.

color{blue} ✍️At other angles, both components are present but one is stronger than the other. There is no stable phase relationship between the two perpendicular components since these are derived from two perpendicular components of an unpolarised beam.

color{blue} ✍️When such light is viewed through a rotating analyser, one sees a maximum and a minimum of intensity but not complete darkness. This kind of light is called partially polarised. Let us try to understand the situation.

color{blue} ✍️When an unpolarised beam of light is incident at the Brewster’s angle on an interface of two media, only part of light with electric field vector perpendicular to the plane of incidence will be reflected.

color{blue} ✍️Now by using a good polariser, if we completely remove all the light with its electric vector perpendicular to the plane of incidence and let this light be incident on the surface of the prism at Brewster’s angle, you will then observe no reflection and there will be total transmission of light.

color{blue} ✍️We began this chapter by pointing out that there are some phenomena which can be explained only by the wave theory. In order to develop a proper understanding, we first described how some phenomena like reflection and refraction, which were studied on this basis of Ray Optics in previously, can also be understood on the basis of Wave Optics.

color{blue} ✍️Then we described Young’s double slit experiment which was a turning point in the study of optics. Finally, we described some associated points such as diffraction, resolution, polarisation, and validity of ray optics. In the next chapter, you will see how new experiments led to new theories at the turn of the century around 1900 A.D.
Q 3128178901

Discuss the intensity of transmitted light when a polaroid sheet is rotated between two crossed polaroids?
Class 12 Chapter 10 Example 8
Solution:

Let I_0 be the intensity of polarised light after passing through the first polariser P_1. Then the intensity of light after passing through second polariser P_2 will be

I = I 0cos^2 θ ,

where θ is the angle between pass axes of p_1 and p_2. Since p_1 and p_3
are crossed the angle between the pass axes of p_2 and p_3 will be
(π//2–θ ). Hence the intensity of light emerging from p_3 will be

I = I_ cos^2 theta cos^2 (pi/2 - theta)

I = I_ cos^2 theta cos^2 theta = (I_0 //4)sin^2theta

Therefore, the transmitted intensity will be maximum when θ = π//4.
Q 3168178905

Unpolarised light is incident on a plane glass surface. What should be the angle of incidence so that the reflected and refracted rays are perpendicular to each other?
Class 12 Chapter 10 Example 9
Solution:

For i + r to be equal to π//2, we should have tan i_B = μ = 1.5. This gives i_B = 57°. This is the Brewster’s angle for air to glass interface.