exemplar-phy-12-wave-optics

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<p><strong>Ques.</strong> Consider a ray of light incident from air onto a slab of glass (refractive index $n$) of width $d$, at an angle $\theta$. The phase difference between the ray reflected by the top surface of the glass and the bottom surface is</p>
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<p>(a) $\frac{4 \pi d}{\lambda} \big(1-\frac{1}{n^{2}} \sin ^{2} \theta\big)^{1 / 2}+\pi$</p>
<p>(b) $\frac{4 \pi d}{\lambda} \big(1-\frac{1}{n^{2}} \sin ^{2} \theta\big)^{1/2}$</p>
<p>(c) $\frac{4 \pi d}{\lambda} \big(1-\frac{1}{n^{2}} \sin ^{2} \theta\big)^{1 / 2}+\frac{\pi}{2}$</p>
<p>(d) $\frac{4 \pi d}{\lambda} \big(1-\frac{1}{n^{2}} \sin ^{2} \theta\big)^{1 / 2}+2 \pi$</p>
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<p><strong>Ans.</strong> (a)</p>
<p>Consider the diagram, the ray $(P)$ is incident at an angle $\theta$ and gets reflected in the direction $P^{\prime}$ and refracted in the direction $P^{\prime \prime}$. Due to reflection from the glass medium, there is a phase change of $\pi$.</p>
<p>Time taken to travel along $O P^{\prime \prime}$</p>
<p>From Snell’s law, $\quad n=\frac{\sin \theta}{\sin r}$</p>
<p>$\Delta t=\frac{O P^{\prime \prime}}{v}=\frac{d / \cos r}{c / n}=\frac{n d}{c \cos r}$</p>
<p><strong><Success>Sol. to be continued …</Success></strong></p>
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<p><strong>Ans.</strong> (a, b, d)</p>
<p>Consider the pattern of the intensity shown in the figure</p>
<p>(i) As intensities of all successive minima is zero, hence we can say that two sources $S_1$ and $S_2$ are having same intensities.</p>
<p>(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.</p>
<p>(iii) We are using monochromatic light in YDSE to avoid overlapping and to have very clear pattern on the screen.</p>
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Polaroid II
<p>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 $90^{\circ}$.</p>
<p><strong><Success>Sol. to be continued …</Success></strong></p>
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<p>Polaroid I<br>
Polaroid II<br>
Consider the above diagram where a third polaroid (III) is placed between polaroid (I) and polaroid II.</p>
<p>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 $90^{\circ}$ to the pass axis of (III).</p>
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<p><strong>Ans.</strong></p>
<p>When angle of incidence is equal to Brewster’s angle, the transmitted light is unpolarised and reflected light is plane polarised.</p>
<p>Consider the diagram in which unpolarised light is represented by dot and plane polarised light is represented by $\to$s.</p>
<p>Polarisation by reflection occurs when the angle of incidence is the Brewster’s angle</p>
<p>i.e.,</p>
<p>$\tan i_B={ }^{1} \mu_2=\frac{\mu_2}{\mu_1} \text{ where } \mu_2<\mu_1$</p>
<p><strong><Success>Sol. to be continued …</Success></strong></p>
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<p><strong>Thinking process :</strong></p>
<p>Resolving power of a microscope is calculated by $\frac{2 \sin \beta}{1.22 \lambda}$, with $\mu$ as refractive index of the medium and $\beta$ is the angle subtented by the objective at the object.</p>
<p><strong>Ans.</strong></p>
<p>We know that</p>
<p>Resolving power = $\frac{1}{d}=\frac{2 \sin \beta}{1.22 \lambda} \Rightarrow d   _{\min }=\frac{1.22 \lambda}{2 \sin \beta}$</p>
<p>where, $\lambda$ is the wavelength of light and $\beta$ is the angle subtended by the objective at the object.</p>
<p><strong><Success>Sol. to be continued …</Success></strong></p>
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<p>Without p $ \qquad A  =A   _{\perp}+A   _{||} $</p>
<p>$A_1=A^{1} \perp+A   _{\perp}^{2}  =A _{\perp}^{0} \sin (k x-\omega t)+A _{\perp}^{0} \sin (k x-\omega t+\phi) $</p>
<p>$A    _{||}  =A   _{||}^{(1)}+A   _{||}^{(2)} $</p>
<p>$A    _{||}  =A   _{||}^{0}[\sin (k x-\omega t)+\sin (k x-\omega t+\phi)] $</p>
<p>where $A   _{\perp}^{0}, A   _{||}^{0}$ are the amplitudes of either of the beam in perpendicular and parallel polarisations.</p>
<p>$\therefore$ Intensity $=\{|A   _{\perp}^{0}|^{2}+|A   _{||}^{0}|^{2}\} [\sin ^{2}(k x-\omega t)(1+\cos ^{2} \varphi+2 \sin \varphi)+\sin ^{2}(k x-\omega t) \sin ^{2} \varphi]$</p>
<p><strong><Success>Sol. to be continued …</Success></strong></p>
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<p><strong>Thinking process :</strong></p>
<p>Whenever a transparent slab of refractive index $\mu$ and thickness $t$ is inserted in the path of the ray the fringes on the screen shifts by $(\mu-1)$ t towards the slab.</p>
<p><strong>Ans.</strong></p>
<p>In case of transparent glass slab of refractive index $\mu$, the path difference will be calculated as $\Delta x=2 d \sin \theta+(\mu-1) L$.</p>
<p>In case of transparent glass slab of refractive index $\mu$,</p>
<p>the path difference $=2 d \sin \theta+(\mu-1) L$.</p>
<p><strong><Success>Sol. to be continued …</Success></strong></p>
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<p>$\therefore \quad 2 d \sin \theta_1+0.5 L  = \pm \frac{\lambda}{2} $</p>
<p>$\text{ or } \quad \sin \theta_1  =\frac{ \pm \lambda / 2-0.5 L}{2 d}=\frac{ \pm \lambda / 2-d / 8}{2 d} $</p>
<p>$ =\frac{ \pm \lambda / 2-\lambda / 8}{2 \lambda}= \pm \frac{1}{4}-\frac{1}{16} $</p>
<p>$[\because$ The diffraction occurs if the wavelength of waves is nearly equal to the side width $(d)]$</p>
<p>On the positive side $\sin \theta_1^{\prime}+=+\frac{1}{4}-\frac{1}{16}=\frac{3}{16}$</p>
<p>On the negative side $\sin \theta^{\prime \prime} _1=-\frac{1}{4}-\frac{1}{16}=-\frac{5}{16}$</p>
<p><strong><Success>Sol. to be continued …</Success></strong></p>
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<p><strong>Ques.</strong>  Four identical monochromatic sources $A, B, C, D$ as shown in the (figure) produce waves of the same wavelength $\lambda$ and are coherent. Two receiver $R_1$ and $R_2$ are at great but equal distances from $B$.</p>
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<p>(i) Which of the two receivers picks up the larger signal?</p>
<p>(ii) Which of the two receivers picks up the larger signal when $B$ is turned off?</p>
<p>(iii) Which of the two receivers picks up the larger signal when $D$ is turned off?</p>
<p>(iv) Which of the two receivers can distinguish which of the sources $B$ or $D$ has been turned off?</p>
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<p>The path difference between the signal from $D$ with that of $A$ is</p>
<p>$ \sqrt{d^{2}+\frac{\lambda}2^{2}}-(d-\lambda / 2)=d \quad 1+{\frac{\lambda}{4 d^{2}}}^{1 / 2}-d+\frac{\lambda}{2} $</p>
<p>$ =d \quad 1+{\frac{\lambda^{2}}{8 d^{2}}}^{1 / 2}-d+\frac{\lambda}{2} \approx \frac{\lambda}{2} \quad(\because d \gg \lambda) $</p>
<p>Therefore, phase difference is $\pi$</p>
<p>$\therefore \quad Y_D=a \cos (\omega t-\pi)=-a \cos \omega t$</p>
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<p><strong><Success>Sol. to be continued …</Success></strong></p>
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<p><strong>Ques.</strong>  The optical properties of a medium are governed by the relative permittivity $(\varepsilon_r)$ and relative permeability $(\mu_r)$. The refractive index is defined as $\sqrt{\mu_r \varepsilon_r}=n$. For ordinary material, $\varepsilon_r>0$ and $\mu_r>0$ and the positive sign is taken for the square root.</p>
<p>In 1964, a Russian scientist V. Veselago postulated the existence of material with $\varepsilon_r<0$ and $\mu_r<0$. Since, then such ‘metamaterials’ have been produced in the laboratories and their optical properties studied. For such materials $n=-\sqrt{\mu_r \varepsilon_r}.$ As light enters a medium of such refractive index the phases travel away from the direction of propagation.</p>
<p>(i) According to the description above show that if rays of light enter such a medium from air (refractive index $=1$) at an angle $\theta$ in 2$^\text{nd}$ quadrant, then the refracted beam is in the 3$^\text{rd}$ quadrant.</p>
<p>(ii) Prove that Snell’s law holds for such a medium.</p>
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<p>(i) Let $A B$ represent the incident wavefront and $D E$ represent the refracted wavefront. All points on a wavefront must be in same phase and in turn, must have the same optical path length.</p>
<p>Thus $\qquad -\sqrt{\varepsilon_r \mu_r} A E  =B C-\sqrt{\varepsilon_r \mu_r} C D $</p>
<p>or $\qquad B C  =\sqrt{\varepsilon_r \mu_r}(C D-A E) $</p>
<p>$B C  >0, C D>A E $</p>
<p>As showing that the postulate is reasonable. If however, the light proceeded in the sense it does for ordinary material (viz. in the fourth quadrant, Fig. 2)</p>
<p><strong><Success>Sol. to be continued …</Success></strong></p>
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<p>Then, $\qquad -\sqrt{\varepsilon_r \mu_r} A E  =B C-\sqrt{\varepsilon_r \mu_r} C D $</p>
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<p>or $\qquad B C  =\sqrt{\varepsilon_r \mu_r}(C D-A E) $</p>
<p>If $B C>0$, then $C D>A E$</p>
<p>which is obvious from Fig (i).</p>
<p>Hence, the postulate reasonable.</p>
<p>However, if the light proceeded in the sense it does for ordinary material, (going from 2$^\text{nd}$ quadrant to 4$^\text{th}$ quadrant) as shown in Fig. (i)., then proceeding as above,</p>
<p><strong><Success>Sol. to be continued …</Success></strong></p>
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<p>Refractive index of film $=1.38$ and refractive index of glass $=1.5$.</p>
<p>Given,</p>
<p>$\lambda=5500 \AA .$</p>
<p>Consider a ray incident at an angle $i$. A part of this ray is reflected from the air-film interface and a part refracted inside.</p>
<p>This is partly reflected at the film-glass interface and a part transmitted. A part of the reflected ray is reflected at the film-air interface and a part transmitted as $r_2$ parallel to $r_1$. Of course successive reflections and transmissions will keep on decreasing the amplitude of the wave.</p>
<p><strong><Success>Sol. to be continued …</Success></strong></p>
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