Physics Ncert Exemplar 2019 Solutions for Class 12 Science Physics Chapter 10 - Wave Optics

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Page No 62:

Question 1:

Consider a light beam incident from air to a glass slab at Brewster’s angle as shown in Fig. 10.1. A polaroid is placed in the path of the emergent ray at point P and rotated about an axis passing through the centre and perpendicular to the plane of the polaroid.

(a) For a particular orientation there shall be darkness as observed through the polaoid.
(b) The intensity of light as seen through the polaroid shall be independent of the rotation.
(c) The intensity of light as seen through the Polaroid shall go through a minimum but not zero for two orientations of the polaroid.
(d) The intensity of light as seen through the polaroid shall go through a minimum for four orientations of the polaroid.

Answer:

Consider the diagram the light beam incident from air to the glass slab at Brewster’s angle (i_p ).
The incident ray is unpolarised and is represented by dot (.).
The reflected light is plane polarised represented by arrows

As the emergent ray is unpolarised, hence intensity cannot be zero when passes through polaroid.

Hence, the correct answer is option C.

Page No 62:

Question 2:

Consider sunlight incident on a slit of width 10⁴ $\overset{o}{A}$ . The image seen through the slit shall

(a) be a fine sharp slit white in colour at the center.
(b) a bright slit white at the center diffusing to zero intensities at the edges.
(c) a bright slit white at the center diffusing to regions of different colours.
(d) only be a diffused slit white in colour.

Answer:

Width of the slit = $10^{4} \overset{o}{A} = 10^{4} \times 10^{- 10} = 10^{- 6} m$ or 10,000 $\overset{o}{A}$ .
The wavelength of (visible) sunlight varies from 4000 $\overset{o}{A}$ to 8000 $\overset{o}{A}$ .
As the width of the slit is comparable to that of wavelength, hence diffraction occurs with maxima at the centre.
So, at the centre all colours appear i.e., mixing of colours form a white patch at the centre.

Hence, the correct answer is option A.

Page No 63:

Question 3:

Consider a ray of light incident from air onto a slab of glass (refractive index n) of width d, at an angle θ. The phase difference between the ray reflected by the top surface of the glass and the bottom surface is

$(a) \frac{4 πd}{λ} {(1 - \frac{1}{n^{2}} \sin^{2} θ)}^{\frac{1}{2}} + π (b) \frac{4 πd}{λ} {(1 - \frac{1}{n^{2}} \sin^{2} θ)}^{\frac{1}{2}} (c) \frac{4 πd}{λ} {(1 - \frac{1}{n^{2}} \sin^{2} θ)}^{\frac{1}{2}} + \frac{π}{2} (d) \frac{4 πd}{λ} {(1 - \frac{1}{n^{2}} \sin^{2} θ)}^{\frac{1}{2}} + 2 π$

Answer:

Consider the diagram, the ray (P) is incident at an angle θ and gets reflected in the direction P′ and refracted in the direction P ''.
Due to reflection from the glass medium, there is a phase change of π.

Time taken to travel along OP ′′
$∆ t = \frac{O P''}{v} = \frac{\frac{d}{\cos r}}{\frac{c}{n}} = \frac{n d}{c \cos r} From Snell' s law, n = \frac{\sin θ}{\sin r} \Rightarrow \sin r = \frac{\sin θ}{n}$
$\cos r = \sqrt{1 - \sin^{2} r} = \sqrt{1 - \frac{\sin^{2} θ}{n^{2}}}$

$∆ t = \frac{n d}{c \sqrt{(1 - \frac{\sin^{2} θ}{n^{2}})}} = \frac{n^{2} d}{c} \sqrt{(1 - \frac{\sin^{2} θ}{n^{2}})}$

Phase difference, $∆ ϕ = \frac{2 π}{T} \times ∆ t = \frac{2 π n d}{λ} {(1 - \frac{\sin^{2} θ}{n^{2}})}^{- \frac{1}{2}}$
So, net phase difference = $∆ ϕ + π$ = $\frac{4 π d}{λ} {(1 - \frac{1}{n^{2}} \sin^{2} θ)}^{- \frac{1}{2}} + π$

Hence, the correct answer is option A.

Page No 63:

Question 4:

In a Young’s double slit experiment, the source is white light. One of the holes is covered by a red filter and another by a blue filter. In this case

(a) there shall be alternate interference patterns of red and blue.
(b) there shall be an interference pattern for red distinct from that for blue.
(c) there shall be no interference fringes.
(d) there shall be an interference pattern for red mixing with one for blue.

Answer:

For the interference pattern to be formed on the screen, the sources should be coherent and emits lights of same frequency and wavelength.

In a Young’s double-slit experiment, when one of the holes is covered by a red filter and another by a blue filter. In this case due to filteration only red and blue lights are present. In YDSE monochromatic light is used for the formation of fringes on the screen.

Thus, in this case there shall be no interference fringes

Hence, the correct answer is option C.

Page No 63:

Question 5:

Figure 10.2 shows a standard two slit arrangement with slits S₁, S₂ . P₁, P₂ are the two minima points on either side of P (Fig. 10.2).
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.

Answer:

According to question, there is a hole at point P₂ .
From Huygen’s principle, wave will propagates from the sources S₁ and S₂ .

Each point on the screen will acts as secondary sources of wavelets. Now, there is a hole at point P₂ (minima). The hole will act as a source of fresh light for the slits S₃ and S₄ .
Therefore, there will be a regular two slit pattern on the second screen.

Hence, the correct answer is option D.

Page No 64:

Question 6:

Two source S₁and S₂ of intensity I₁ and I₂ are placed in front of a screen [Fig. 10.3 (a)]. The patteren of intensity distribution seen in the central portion is given by Fig. 10.3 (b).

In this case which of the following statements are true.

(a) S₁ and S₂have the same intensities.
(b) S₁ and S₂ have a constant phase difference.
(c) S₁ and S₂ have the same phase.

Answer:

Consider the pattern of the intensity shown in the question figure
(i) As intensities of all successive minima is zero, hence we can say that two sources S₁ and S₂ are having same intensities.
(ii) As width of the successive maxima (pulses) increases in continuous manner, we can say that the path difference (x) or phase difference varies in continuous manner.
(iii) We are using monochromatic light in YDSE to avoid overlapping and to have very clear pattern on the screen.

Hence, the correct answer are option A, B and D.

Page No 64:

Question 7:

Consider sunlight incident on a pinhole of width 10³ $\overset{o}{A}$ . The image of the pinhole seen on a screen shall be

(a) a sharp white ring.
(b) different from a geometrical image.
(c) a diffused central spot, white in colour.
(d) diffused coloured region around a sharp central white spot.

Answer:

Diffraction: It is the phenomenon of the bending of light around small obstacles.
The size of obstacle must be in comparable size with wavelength of the light or $d \leq λ$ .
Width of the pinhole = $1000 \overset{o}{A}$
The wavelength of sunlight ranges from 4000 Å to 8000 Å.

Clearly, width of the slit $\leq$ wavelength of the sunlight.

Thus, light is diffracted from the hole.
Due to diffraction from the light the image formed on the screen will be different from the geometrical image.

Hence, the correct answer is option B and D.

Page No 64:

Question 8:

Consider the diffraction pattern for a small pinhole. As the size of the hole is increased

(a) the size decreases.
(b) the intensity increases.
(c) the size increases.
(d) the intensity decreases.

Answer:

The width of central maxima, $β_{o} = λ \frac{D}{d}$
Width of secondary maxima, $\frac{λ}{d}$

So on increasing width of hole of pinhole, d increases. Thus, the size of central maxima decreases.

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As the energy passing through hole increased on increasing the size of hole. So the intensity of pattern will increase.

Hence, the correct answers are option A and B.

Page No 64:

Question 9:

For light diverging from a point source

(a) the wavefront is spherical.
(b) the intensity decreases in proportion to the distance squared.
(c) the wavefront is parabolic.
(d) the intensity at the wavefront does not depend on the distance.

Answer:

Consider the diagram in which light diverges from a point source (O).

Due to the point source light propagates in all directions symmetrically and hence, wavefront will be spherical as shown in the diagram.

If power of the source is P, then intensity of the source will be
$I = \frac{P}{4 π r^{2}}$
where, r is radius of the wavefront at any time.

Hence, the correct answer is option A and B.

Page No 64:

Question 10:

Is Huygen’s principle valid for longitudinal sound waves?

Answer:

When we are considering a point source of sound wave. The disturbance due to the source propagates in spherical symmetry that is in all directions. The formation of wavefront is in accordance with Huygen’s principle.

So, Huygen’s principle is valid for longitudinal sound waves also.

Page No 64:

Question 11:

Consider a point at the focal point of a convergent lens. Another convergent lens of short focal length is placed on the other side. What is the nature of the wavefronts emerging from the final image?

Answer:

Consider the ray diagram shown below

The point image I₁ , due to L₁ is at the focal point.
Now, due to the converging lense L₂ , let final image formed is I which is point image, hence the wavefront for this image will be of spherical symmetry.

Page No 64:

Question 12:

What is the shape of the wavefront on earth for sunlight?

Answer:

Due to the large distance the radius of wavefront can be considered as large (infinity) and hence, wavefront is almost plane.

Page No 65:

Question 13:

Why is the diffraction of sound waves more evident in daily experience than that of light wave?

Answer:

The necessary condition for diffraction to happen is that the width of the obstacle must be less than or comparable to the wavelength of the wave.

As we know that the frequencies of sound waves lie between 20 Hz to 20 kHz so that their wavelength ranges between 15 m to 15 mm. The diffraction occur if the wavelength of waves is nearly equal to slit width.

As the wavelength of light waves is 7000 $\times 10^{- 10}$ to 4000 $\times 10^{- 10}$ m. The slit width is very near to the wavelength of sound waves as compared to light waves. Thus, the diffraction of sound waves is more evident in daily life than that of light waves.

Page No 65:

Question 14:

The human eye has an approximate angular resolution of $ϕ$ = 5.8 × 10^–4 rad and a typical photoprinter 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.

Answer:

Angular dispersion of human eye, $ϕ = 5.8 \times 10^{- 4} rad$
Number of dots printed by printer in one inch = 300
The linear distance between the two dots, $l = \frac{2.54}{300} cm = 0.84 \times 10^{- 2} cm .$
At a distance of z cm, this subtends an angle, $ϕ = \frac{l}{z}$
$\Rightarrow z = \frac{l}{ϕ} = 14.5 cm$

Page No 65:

Question 15:

A polariod (I) is placed in front of a monochromatic source. Another polatiod (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.

Answer:

In the diagram shown, a monochromatic light is placed infront of polaroid (I) as shown below.

As per the given question, monochromatic light emerging from polaroid (I) is plane polarised. When polaroid (II) is placed infront of this polaroid (I), and rotated till no light passes through polaroid (II), then (I) and (II) are set in crossed positions, i.e., pass axes of I and II are at 90°.

Consider the above diagram where a third polaroid (III) is placed between polaroid (I) and polaroid II.

When a third polaroid (III) is placed in between (I) and (II), no light will emerge from (II), if pass axis of (III) is parallel to pass axis of (I) or (II). In all other cases, light will emerge from (II), as pass axis of (II) will no longer be at 90° to the pass axis of (III).

Page No 65:

Question 16:

Can reflection result in plane polarised light if the light is incident on the interface from the side with higher refractive index?

Answer:

When angle of incidence is equal to Brewster’s angle, the transmitted light is unpolarised and reflected light is plane polarised.

Consider the diagram in which unpolarised light is represented by dot and plane polarised light is represented by arrows.
Polarisation by reflection occurs when the angle of incidence is the Brewster’s angle, $\tan i_{B} =^{1} μ_{2} = \frac{μ_{2}}{μ_{1}}$
when the light rays travels in such a medium, the critical angle is, $\sin i_{c} = \frac{μ_{2}}{μ_{1}}$
Where, $μ_{2} < μ_{1}$

As $|\tan i_{B}| > |\sin i_{c}| for large angle i_{B} < i_{C}$
Thus, the polarisation by reflection occurs definitely.

Page No 65:

Question 17:

For the same objective, find the ratio of the least separation between two points to be distinguished by a microscope for light
of 5000 $\overset{\circ}{A}$ and electrons accelerated through 100V used as the illuminating substance.

Answer:

If, λ is the wavelength of light and β is the angle subtended by the objective at the object
Resolving power = $\frac{1}{d} = \frac{2 \sin β}{1.22 λ} \Rightarrow d_{m i n} = \frac{1.22 λ}{2 \sin β}$

Here, $λ = 5500 \overset{o}{A}$
Then, $d_{m i n} = \frac{1.22 \times 5500 \times 10^{- 10}}{2 \sin β}$

For electrons accelerated through 100 V, the de-Broglie wavelength, $λ = \frac{12.27}{\sqrt{V}} = \frac{12.27}{\sqrt{100}} = 0.12 \times 10^{- 9} m$
Thus, $d'_{m i n} = \frac{1.22 \times 0.12 \times 10^{- 9} \times 10^{- 19}}{2 \sin β}$

Taking ratio, $\frac{d'_{m i n}}{d_{m i n}} = \frac{0.12 \times 10^{- 9}}{5500 \times 10^{- 10}} = 0.2 \times 10^{- 3}$

Page No 65:

Question 18:

Consider a two slit interference arrangements (Fig. 10.4) 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.

Answer:

Taking reference from the given figure, we can write:
$T_{2} P = T_{2} O + O P = D + X and T_{1} P = T_{1} O - O P = D - x S_{1} P = \sqrt{{(S_{1} T_{1})}^{2} + {(P T_{1})}^{2}} = \sqrt{D^{2} + {(D - x)}^{2}} and S_{2} P = \sqrt{{(S_{2} T_{2})}^{2} + {(P T_{2})}^{2}} = \sqrt{D^{2} + {(D + x)}^{2}} For minima, S_{2} P - S_{1} P = (2 n - 1) \frac{λ}{2} For first minima, n = 1, \sqrt{D^{2} + {(D + x)}^{2}} - \sqrt{D^{2} + {(D - x)}^{2}} = \frac{λ}{2} If x = D Then, \sqrt{D^{2} + 4 D^{2}} - \sqrt{D^{2} + 0^{2}} = \frac{λ}{2} Solving, the above equation D = 0.404 λ$

Page No 65:

Question 19:

Figure 10.5 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 polariser is present, calculate in the present case, the intensity of the principal maxima as well as of the first minima.

Answer:

Let $A_{⊥}$ and A_|| be the amplitude of the either beam of light in perpendicular (⊥) and parallel (||) polarisation.
(A¹) and (A²) indicates the amplitude from the first and second slit.
Now, intensity calculation without the presence of P in the path:
$A = A_{⊥} + A_{| |} A_{⊥} = A_{⊥}^{1} + A_{⊥}^{2} = A_{⊥}^{0} \sin (k x - ω t) + A_{⊥}^{0} \sin (k x - ω t + ϕ)$
(Let Ï• be the phase difference)
$A_{| |} = A_{| |}^{1} + A_{| |}^{2} = A_{| |}^{0} \sin (k x - ω t) + A_{| |}^{0} \sin (k x - ω t + ϕ)$
Now, intensity,
$I = {(A_{⊥}^{1} + A_{⊥}^{2})}^{2} + {(A_{| |}^{1} + A_{| |}^{2})}^{2} \Rightarrow I = [{|A_{⊥}^{0}|}^{2} + {|A_{| |}^{0}|}^{2}] (\frac{1}{2}) \times 2 (1 + \cos ϕ) \Rightarrow I = 2 {|A_{⊥}^{0}|}^{2} (1 + \cos ϕ) [∵ {|A_{⊥}^{0}|}_{average} = {|A_{| |}^{0}|}_{average}]$
When no polariser is present, then intensity at principal maxima, $I_{o} = 2 {|A_{⟂}^{0}|}^{2} \times 2 = 4 {|A_{⟂}^{0}|}^{2} . . . . . (1)$
Now, intensity calculation, when P is present in the path as shown in the diagram in the given question:
When, P is present then, $A_{⊥}^{2}$ is blocked:
So, intensity, $I^{'} = {(A_{⊥}^{1})}^{2} + {(A_{| |}^{1} + A_{| |}^{2})}^{2}$
$= 2 {|A_{⊥}^{0}|}^{2} (\cos^{2} \frac{ϕ}{2}) + {|A_{⊥}^{0}|}^{2} \frac{1}{2}$
So, intensity at principal maxima, when polariser is present:
$\begin{array}{rcl} I^{'} & = & {|A_{⊥}^{0}|}^{2} (2 + \frac{1}{2}) \\ = & \frac{I_{o}}{4} \times \frac{5}{2} [As, I_{o} = 4 {|A_{⊥}^{0}|}^{2}] \\ = & \frac{5}{8} I_{o} \end{array}$
Now, intensity at first minima with polariser:
$\begin{array}{rcl} I^{''} & = & 0 + {|A_{⊥}^{0}|}^{2} \times \frac{1}{2} \\ = & 0 + \frac{I_{o}}{4} \times \frac{1}{2} = \frac{I_{o}}{8} \end{array}$

Page No 66:

Question 20:

AC = CO = D, S₁C = S₂C = d << D

A small transparent slab containing material of μ =1.5 is placed along AS₂(Fig.10.6). 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.

Answer:

Given, μ = 1.5, L = $\frac{d}{4}$ for the slab.
Slit width = 2d and screen distance = D,
Path difference,
$({AS}_{2} + S_{2} P_{1}) - ({AS}_{1} + S_{1} P_{1}) ∆ x = 2 d sinθ + (μ - 1) L [Assume, d < < D] \Rightarrow ∆ x = 2 d sinθ + (1.5 - 1) \frac{d}{4} \Rightarrow ∆ x = 2 d sinθ + 0.5 \frac{d}{4} = 2 d sinθ + \frac{d}{8} . . . . . (1)$
Now from given diagram in the question:
CO = D
$\tan θ = \frac{{OP}_{1}}{CO} \Rightarrow {OP}_{1} = CO \tan θ = D \tan θ$
So, for principal maxima, path difference = 0
$\Rightarrow 0 = 2 d \sin θ_{o} + \frac{d}{8} (Let θ = θ_{o})$
$\Rightarrow \sin θ_{o} = - \frac{1}{16} So, {OP}_{1} = D \tan θ_{o} \approx D \sin θ_{o} \approx - \frac{D}{16}$
Here, θ_o is the angular position corresponding to the principal maxima.
But for the first maxima, the path difference = $\pm \frac{λ}{2}$
So, $∆ x = \pm \frac{λ}{2} = 2 d \sin θ_{m} + \frac{d}{8} [from equation (1)]$
(Let θ_m be the angular position of first maxima)
$\Rightarrow \sin θ_{m} = \frac{\pm \frac{λ}{2} - \frac{d}{8}}{2 d}$
[âˆµ diffraction occurs if the wavelength (λ) of the wave is nearly equal to d]
$\Rightarrow \sin θ_{m} = \frac{\pm \frac{λ}{2} - \frac{λ}{8}}{2 λ} = \pm \frac{1}{4} - \frac{1}{16} \Rightarrow \sin θ_{m}^{+} = \frac{3}{16} (for + \frac{λ}{2}) and \Rightarrow \sin θ_{m}^{-} = - \frac{5}{16} (for - \frac{λ}{2})$
So, the position of first principal maxima on positive side from point O,
$D \tan θ_{m}^{+} = D \frac{\sin θ_{m}^{+}}{\cos θ_{m}^{+}} = D \frac{\sin θ_{m}^{+}}{\sqrt{1 - \sin^{2} θ_{m}^{+}}} = \frac{3 D}{\sqrt{247}}$
Also, on the negative side is at a distance,
$\begin{array}{rcl} D \tan θ_{m}^{-} & = & D \frac{\sin θ_{m}^{-}}{\sqrt{1 - \sin^{2} θ_{m}^{-}}} \\ = & D \frac{(- \frac{5}{16})}{\sqrt{1 - {(\frac{5}{16})}^{2}}} \\ = & \frac{- 5 D}{\sqrt{231}} below the point O . \end{array}$

Page No 66:

Question 21:

Four identical monochromatic sources A,B,C,D as shown in the (Fig.10.7) produce waves of the same wavelength λ and are coherent. Two receiver R₁ and R₂ are at great but equal distaces 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?

Answer:

(i) Let the wave from A at R₁ be,
y_A = a cos ωt .....(1) (a = amplitude, ω = angular frequency)
Phase difference = Path difference × $(\frac{2 π}{λ})$
So, phase difference of signal reached to R₁ from A and B,
$∆ ϕ_{AB} = A B \times \frac{2 π}{λ} = \frac{λ}{2} \times \frac{2 π}{λ} = π$
Thus, the wave equation at R₁ from source at B will be,
$y_{B} = a \cos (ω t - ϕ_{AB}) = a \cos (ω t - π) \Rightarrow y_{B} = - a \cos ω t . . . . . (2)$
Similarly, phase difference of the signal from C with that from A,
$ϕ_{AC} = A C \times \frac{2 π}{λ} = (\frac{λ}{2} + \frac{λ}{2}) \times \frac{2 π}{λ} = 2 π$
So, the wave equation at R₁ due to C will be:
$y_{C} = a \cos (ω t - ϕ_{AC}) = a \cos (ω t - 2 π) = a \cos ω t . . . . . (3)$
Now, the phase difference between the signal from D with that of A at point R₁ will be,
$ϕ_{AD} = A D \times \frac{2 π}{λ} ϕ_{AD} = [\sqrt{d^{2} + {(\frac{λ}{2})}^{2}} - (d - \frac{λ}{2})] \times \frac{2 π}{λ}$
Using the approximation, d >>> λ
$\Rightarrow ϕ_{AD} \approx (\frac{λ}{2}) \times \frac{2 π}{λ} \Rightarrow ϕ_{AD} \approx π$
Thus, $y_{D} = a \cos (ω t - ϕ_{AD}) = a \cos (ω t - π) = - a \cos (ω t) . . . . . (4)$
Hence, the signal picked up at R₁ due to source A, B, C and D will be,
$y = y_{A} + y_{B} + y_{C} + y_{D} \Rightarrow y = a \cos ω t - a \cos ω t + a \cos ω t - a \cos ω t \Rightarrow y = 0$
Similarly, for receiver R₂:
Let, wave signal at R₂ from B be,
$y_{B}^{'} = a_{1} \cos (ω t) . . . . . (5) y_{D}^{'} = - a_{1} \cos (ω t) . . . . . (6)$
Path difference between signal at A and B will be,
$= \sqrt{d^{2} + {(\frac{λ}{2})}^{2}} - d = d \sqrt{1 + \frac{λ^{2}}{4 d^{2}}} - d \approx \frac{λ^{2}}{8 d^{2}}$
So, phase difference,
$ϕ_{AB} = \frac{λ^{2}}{8 d^{2}} \times \frac{2 π}{λ} = \frac{πλ}{4 d}$
So, the wave at R₂ due to A will be,
$y_{A}^{'} = a_{1} \cos (ω t - ϕ_{AB})$
Similarly,
$y_{C}^{'} = a_{1} \cos (ω t - ϕ_{AC})$
Let, $ϕ_{AB} = ϕ_{AC} = ϕ$
The, signal picked up by R₂ is,
$y^{'} = y_{A}^{'} + y_{B}^{'} + y_{C}^{'} + y_{D}^{'} y^{'} = a_{1} \cos (ω t - ϕ) + a_{1} \cos (ω t) + a_{1} \cos (ω t - ϕ) - a_{1} \cos (ω t) = 2 a_{1} \cos (ω t - ϕ) {|y^{'}|}^{2} = 4 a_{1}^{2} \cos^{2} (ω t - ϕ) <I_{R_{2}}> = 2 a_{1}^{2}$
Hence, the receiver R₂ will pick up larger signal.
(ii) Now, if source B is switched off:
So, R₁ will pick up:
y = a cos ωt
$<I_{R_{1}}>$ = a²2ωt>
$<I_{R_{1}}> = \frac{a^{2}}{2}$
Also, R₂ picks up, y = a cos ωt
$\begin{array}{rcl} <I_{R_{2}}> & = & a^{2} < \cos^{2} ω t > \\ = & \frac{a^{2}}{2} \end{array}$
Hence, both the receivers pick up same signal.
(iii) Now, if D is switched off then, receiver R₁will pick up
y = a cos ωt
So, $<I_{R_{1}}> = \frac{a^{2}}{2}$
and receiver R₂ will pick up
y = 3a cos ωt
$<I_{R_{2}}> = \frac{9 a^{2}}{2}$
Hence, R₂ will pick up larger signal than R₁.
(iv) Signal at receiver R₁ indicates that B has been switched off and an enhanced signal at R₂ indicates that D has been switched off.

Page No 67:

Question 22:

The optical properties of a medium are governed by the relative permitivity (ε_r) and relative permeability (μ_r). The refractive index is defined as $\sqrt{μ_{r} ε_{r}}$ = n.For ordinary material ε_r > 0 and μ_r > 0 and the positive sign is taken for the square root. In 1964, a Russian
scientist V. Veselago postulated the existence of material with ε_r < 0 and μ_r< 0. Since then such ‘metamaterials’ have been produced in the laboratories and their optical properties studied. For such materials n = $- \sqrt{μ_{r} ε_{r}}$ . 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 air (refractive index =1) at an angle θ in 2^ndquadrant, them the refracted beam is in the 3rd quadrant.
(ii) Prove that Snell’s law holds for such a medium.

Answer:

(i) Given refractive index of the medium, n = $\sqrt{μ_{r} ε_{r}}$     (μ_r> 0 and ε_r > 0).
But for meta-materials,
n = $- \sqrt{μ_{r} ε_{r}}$    (μ_r< 0 and ε_r < 0).
So, if the ray of light enters a medium (n < 0) at an angle θ_i as show in diagram:

Fig (1)
Here, ED shows the wavefront, i.e., all the points on this must have same phase.
So, according to Vaselago postulate,
$- \sqrt{μ_{r} ε_{r}} A E = B C - \sqrt{ε_{r} μ_{r}} CD [Optical path length]$
$\Rightarrow BC = \sqrt{ε_{r} μ_{r}} (CD - AE)$
As shown in the diagram
BC > 0 and CD > AE
So, the incident ray is in 2nd quadrant, the refracted ray is in 3rd quadrant.
Hence, the postulate is reasonable.
Now, if the ray emerges out in the fourth quadrant as shown in diagram below:

                            Fig (2)
$- \sqrt{ε_{r} μ_{r}} AE = BC - \sqrt{ε_{r} μ_{r}} CD \Rightarrow BC = \sqrt{ε_{r} μ_{r}} (CD - AE)$
In this case,
AE > CD
⇒ CD – AE < 0
⇒ BC < 0
which is not possible.
Hence, the postulate is correct.
(ii) For Figure (1)
BC = AC sin θ_i
And, CD – AE = AC sin θ_r
Since, $- \sqrt{ε_{r} μ_{r}} (AE - CD) = BC$
Dividing both side by AC we get:
$\Rightarrow \frac{BC}{AC} = (- \sqrt{ε_{r} μ_{r}}) \frac{(AE - CD)}{AC} \Rightarrow {sinθ}_{i} = (- \sqrt{ε_{r} μ_{r}}) \times {sinθ}_{r} \Rightarrow 1 \times {sinθ}_{i} = - n {sinθ}_{r} (Snell' s law)$
⇒ Snell's law holds good for such medium.

Page No 67:

Question 23:

To ensure almost 100 per cent transmittivity, 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 center of the visible speetrum (5500 $\overset{°}{A}$ ) there is maximum transmission.

Answer:

Ray incident on the air-film interface and get refracted at an angle r But, it gets totally internally reflection from the film-glass interface and after reflection at D it get refraction at film-air interface again at C.
Let angle of incidence by i and angle of refraction be r as shown in the diagram.
The refractive index of the medium,
n₁ = 1 (air), n₂ = 1.38 (film), n₃ = 1.5 (glass)
Let thickness of the prism be d.
So the optical path difference,
n (AD + CD) – AB
But, AD = CD = $\frac{d}{\cos r}$

$[In the shown image, AD = CD, AP = MD = d] [In triangle AMD, \cos r = \frac{MD}{AD} = \frac{d}{AD}]$
Similarly,
AB = AC sini
$\frac{A C}{2} = d \tan r$
So, AC = 2d tanr
⇒ AB = 2d tanr sini
Hence, the optical path difference
$= 2 n_{2} \frac{d}{\cos r} - 2 d \tan r \sin i = 2 n_{2} d \cos r (∵ \tan r = \frac{\sin r}{\cos r})$
But for distructive interference path difference $= \frac{λ}{2}$
$\Rightarrow 2 n_{2} d \cos r = \frac{λ}{2} \Rightarrow n_{2} d \cos r = \frac{λ}{4}$
Now, very very small angle of incidence,
$i \approx r \approx 0 So, n_{2} d \approx \frac{λ}{4} i f λ = 5500 \overset{°}{A}, \Rightarrow d = \frac{5500 \overset{°}{A}}{1.38 \times 4} \approx 996 \overset{°}{A}$

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