Wave Optics

The figure shown a two-slit arrangement with a source which emits unpolarised light. P is a polariser with axis whose direction is not given. If I0 is the intensity of the principal maxima when no polarizer is present, calculate in the present case, the intensity of the principal maxima as well as of the first minima.

Given wave amplitude in perpendicular polarisation $ A_{\perp}=A_{\perp}^{0}(\sin (k x-\omega t)+\sin (k x-\omega t+\phi)) $ wave amplitude in parallel polarisation $ A_{\|}=A_{\|}^{0}(\sin (k...

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AC = CO = D, S1 C = S2 C = d << D A small transparent slab containing material of µ =1.5 is placed along AS2. What will be the distance from O of the principal maxima and of the first minima on either side of the principal maxima obtained in the absence of the glass slab?

According to the question, $ \begin{array}{l} \Delta x=2 d \sin \theta+(\mu-1) L \\ \sin \theta 0=-1 / 16 \end{array} $ From central maxima, $O P=-D / 16$ $ \sin \theta_{1}=\frac{\pm \lambda / 2-d /...

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The figure shows a standard two-slit arrangement with slits S1, S2. P1, P2 are the two minima points on either side of P. At P2 on the screen, there is a hole and behind P2 is a second 2- slit arrangement with slits S3, S4 and a second screen behind them;(a) There would be no interference pattern on the second screen but it would be lighted (b) The second screen would be totally dark (c) There would be a single bright point on the second screen (d) There would be a regular two-slit pattern on the second screen

The correct answer is d) there would be a regular two-slit pattern on the second screen

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The human eye has an approximate angular resolution of φ = 5.8 × 10–4 rad and a typical photo printer prints a minimum of 300 dpi (dots per inch, 1 inch = 2.54 cm). At what minimal distance z should a printed page be held so that one does not see the individual dots.

Angular separation, φ = 5.8 × 10–4 rad The average distance between the two dots = 2.54/300 = 0.85 × 10-2 cm At distance z cm, the angle subtended = 0.85 × 10-2/z Resolution angle for human = 0.85 ×...

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A polaroid (I) is placed in front of a monochromatic source. Another polaroid (II) is placed in front of this polaroid (I) and rotated till no light passes. A third polaroid (III) is now placed in between (I) and (II). In this case, will light emerge from (II)? Explain.

A monochromatic source is placed in front of polaroid (I), and polaroid (II) is positioned in front of polaroid (I) (I). The light is unaltered as it travels through the polaroid (II). The Polaroid...

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Consider a two-slit interference arrangements such that the distance of the screen from the slits is half the distance between the slits. Obtain the value of D in terms of λ such that the first minima on the screen falls at a distance D from the centre O.

The minima will occur when ∆x = S2P – S1P = (2n-1/2)λ S1P = √D2 + (D – x)2 S2P = √D2 + (D + x)2 T2P = D + x T1P = D – x [D2 + (D+x)2]-1/2 – [D2+(D-x)2]1/2 = λ/2   D =...

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Four identical monochromatic sources A, B, C, D as shown in the figure produce waves of the same wavelength λ and are coherent. Two receiver R1 and R2 are at great but equal distances from B. (i) Which of the two receivers picks up the larger signal? (ii) Which of the two receivers picks up the larger signal when B is turned off? (iii) Which of the two receivers picks up the larger signal when D is turned off? (iv) Which of the two receivers can distinguish which of the sources B or D has been turned off?

i) R2 is the larger signal. ii) When B is switched off, R1 is y = a cos ꞷt IR1 = a2/2 R2 is y = a cos ꞷt IR2 = a2/2 So, it can be said that R1 and R2 are the same. iii) When D is switched off, R1 is...

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The optical properties of a medium are governed by the relative permittivity (εr) and relative permeability (µr ). The refractive index is defined as √µr εr = n. For ordinary material εr > 0 and µr > 0 and the positive sign is taken for the square root. In 1964, a Russian scientist V. Veselago postulated the existence of material with εr < 0 and µr < 0. Since then such ‘metamaterials’ have been produced in the laboratories and their optical properties studied. For such materials n = -√µrεr. As light enters a medium of such refractive index the phases travel away from the direction of propagation. (i) According to the description above show that if rays of light enter such a medium from the air (refractive index =1) at an angle θ in 2nd quadrant, them the refracted beam is in the 3rd quadrant (ii) Prove that Snell’s law holds for such a medium

BC = √μrεr (CD-AE) BC >0; CD > AE -√μrεr AE = BC – √μrεr CD BC = √μrεr (CD – AE) AE > CD Therefore, BC < 0 ii) BC = AC sin θi CD – AE = AC sin θr BC = √μrεr AC sin θi = √μrεr AC sin θr...

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To ensure almost 100 per cent transmissivity, photographic lenses are often coated with a thin layer of dielectric material. The refractive index of this material is intermediated between that of air and glass (which makes the optical element of the lens). A typically used dielectric film is MgF2 (n = 1.38). What should the thickness of the film be so that at the centre of the visible spectrum (5500 Ao) there is maximum transmission?

IA is incident at point A, forming an incident angle I between the air and the film surface. The reflected and refracted rays are AR1 and AD, respectively. D is the point where the partial...

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Answer the following questions:
(a) When a low flying aircraft passes overhead, we sometimes notice a slight shaking of the picture on our TV screen. Suggest a possible explanation.
(b) As you have learnt in the text, the principle of linear superposition of wave displacement is basic to understanding intensity distributions in diffraction and interference patterns. What is the justification of this principle?

Answer – (a) The aircraft's weak radar emissions interfere with the antenna's reception of the TV signal. (b) This is because the linear aspect of a differential equation that drives wave motion...

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A parallel beam of light of wavelength 500 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1 m away. It is observed that the first minimum is at a distance of 2.5 mm from the centre of the screen. Find the width of the slit.

Answer – We are given that, λ = 500 nm is the wavelength of the beam of light D= 1 m is the distance between the slit and the screen x = 2.5 mm = 2.5 x 10-3 m is the distance of the first...

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Answer the following questions:
(a) In a single slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the central diffraction band?
(b) In what way is diffraction from each slit related to the interference pattern in a double-slit experiment?
(c) When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the centre of the shadow of the obstacle. Explain why?
(d) Two students are separated by a 7 m partition wall in a room 10 m high. If both light and sound waves can bend around obstacles, how is it that the students are unable to see each other even though they can converse easily?
(e) Ray optics is based on the assumption that light travels in a straight line. Diffraction effects (observed when light propagates through small apertures/slits or around small obstacles) disprove this assumption. Yet the ray optics assumption is so commonly used in an understanding location and several other properties of images in optical instruments. What is the justification?

Answer – (a) In a single slit diffraction experiment, doubling the slit width reduces the size of the central diffraction band by half while increasing the intensity of the band by four times. (b)...

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For sound waves, the Doppler formula for frequency shift differs slightly between the two situations: (i) source at rest; observer moving, and (ii) source moving; observer at rest. The exact Doppler formulas for the case of light waves in vacuum are, however, strictly identical for these situations. Explain why this should be so. Would
you expect the formulas to be strictly identical for the two situations in the case of light travelling in a medium?

Answer – Because sound waves can only move through the medium, the Doppler formula differs somewhat between the two scenarios. In both circumstances, the observer's motion relative to the medium is...

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Let us list some of the factors, which could possibly influence the
speed of wave propagation:
(i) nature of the source.
(ii) the direction of propagation.
(iii) the motion of the source and/or observer.
(iv) wavelength.
(v) the intensity of the wave. On which of these factors, if any, does (a) the speed of light in a vacuum,
(b) the speed of light in a medium (say, glass or water), depend?

Answer – (a) In the vacuum, none of the components listed affect the speed of light. (b) The wavelength of light in a media affects the speed of light in that medium.

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You have learnt in the text how Huygens’ principle leads to the laws of reflection and refraction. Use the same principle to deduce directly that a point object placed in front of a plane mirror produces a virtual image whose distance from the mirror is equal to the object distance from the mirror.

Answer – Consider an item O placed at a distance r in front of the plane mirror MO'. The object is designated as point O, and a circle is drawn around it, just touching the plane mirror at point O'....

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Explain how Corpuscular theory predicts the speed of light in a medium, say, water, to be greater than the speed of light in vacuum. Is the prediction confirmed by experimental determination of the speed of light in water? If not, which alternative picture of light is consistent with experiment?

Answer – The velocity of light in the denser medium (water) is greater than the velocity of light in the rarer medium (air), according to Newton's Corpuscular hypothesis (vacuum). This was erroneous...

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In a double-slit experiment, 0.2° is found to be the angular width of a fringe on a screen placed 1 m away. The wavelength of light used is 600 nm. What will be the angular width of the fringe if the entire experimental apparatus is immersed in water? Take refractive index of water to be ¾.

Answer: We are given, Distance of the screen from the slits is D = 1m Wavelength of light used is λ1​ = 600 nm Angular width of the fringe in air is θ1​ = 0.2° θ2​ is the angular...

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In Young’s double-slit experiment, 0.28mm separation between the slits and the screen is placed 1.4m away. 1.2cm is the distance between the central bright fringe and the fourth bright fringe. Determine the wavelength of light used in the experiment.

Answer – We are given, Distance between the screen and the slits, D = 1.4 m Distance between the slits is d = 0.28 mm = 0.28 x 10-3 m Also, the distance between the central fringe and the...

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