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NCERT Exemplar · Class 12 Physics Wave Optics

23 questions

Multiple Choice Questions (Type I) 1–5 (part 1 of 5)

  1. Exercise 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.
    NCERT_Question_Class12_Physics_Exemplar_Ch10_Q1
    NCERT’s answer
    (c)
    (c) Only the reflected ray is fully polarised; the emergent ray is partially polarised, so the minimum is non-zero, twice per turn.\[\tan\theta_B=n \]\[I(\varphi)=I_\parallel\cos^2\varphi+I_\perp\sin^2\varphi,\qquad I_\parallel>I_\perp>0 \]\[I_{\max}=I_\parallel\ \ (\varphi=0^\circ,\,180^\circ),\qquad I_{\min}=I_\perp\neq0\ \ (\varphi=90^\circ,\,270^\circ) \]NCERT_Solution_Class12_Physics_Exemplar_Ch10_Q1\(\displaystyle \varphi\): polaroid axis from the plane of incidence; \(\displaystyle I_\perp<I_\parallel\) as reflection removes only \(\displaystyle E_\perp\).

  2. Exercise 2

    Consider sunlight incident on a slit of width 104 A\displaystyle 10^4 \mathrm{~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.
    (c) The slit width is comparable to visible wavelengths, so diffraction is strong. Every colour peaks at the centre (white); the minima of different colours fall at different angles.\[a=10^{4}\,\mathrm{\AA}=10^{-6}\,\mathrm{m},\qquad 4000\,\mathrm{\AA}\le\lambda\le7000\,\mathrm{\AA} \]\[a\sin\theta_1=\lambda\ \Rightarrow\ \sin\theta_1=\frac{4000}{10^{4}}=0.40\ \text{(violet)},\qquad \frac{7000}{10^{4}}=0.70\ \text{(red)} \]\[\theta=0:\ \text{all colours in phase}\ \Rightarrow\ \text{white central maximum} \]NCERT_Solution_Class12_Physics_Exemplar_Ch10_Q2

  3. Exercise 3

    Consider a ray of light incident from air onto a slab of glass (refractive index n\displaystyle n) of width d\displaystyle d, at an angle θ\displaystyle \theta. The phase difference between the ray reflected by the top surface of the glass and the bottom surface is
    (a)
    4πdλ(1−1n2sin⁡2θ)1/2+π\displaystyle \frac{4 \pi d}{\lambda}\left(1-\frac{1}{n^2} \sin ^2 \theta\right)^{1 / 2}+\pi
    (b)
    4πdλ(1−1n2sin⁡2θ)1/2\displaystyle \frac{4 \pi d}{\lambda}\left(1-\frac{1}{n^2} \sin ^2 \theta\right)^{1 / 2}
    (c)
    4πdλ(1−1n2sin⁡2θ)1/2+π2\displaystyle \frac{4 \pi d}{\lambda}\left(1-\frac{1}{n^2} \sin ^2 \theta\right)^{1 / 2}+\frac{\pi}{2}
    (d)
    4πdλ(1−1n2sin⁡2θ)1/2+2π\displaystyle \frac{4 \pi d}{\lambda}\left(1-\frac{1}{n^2} \sin ^2 \theta\right)^{1 / 2}+2 \pi.
    NCERT’s answer
    (a)
    (a) The top reflection (air to glass) gains an extra \(\displaystyle \pi\); the bottom reflection (glass to air) gains none. Here \(\displaystyle \lambda\) is the wavelength inside the glass.NCERT_Solution_Class12_Physics_Exemplar_Ch10_Q3\[\sin\theta=n\sin r\ \Rightarrow\ \cos r=\left(1-\frac{1}{n^2}\sin^2\theta\right)^{1/2} \]\[n\,(AB+BC)-AD=\frac{2nd}{\cos r}-2d\tan r\,\sin\theta=2nd\cos r \]\[\delta=\frac{2\pi}{\lambda_0}\,(2nd\cos r)+\pi=\frac{4\pi d}{\lambda}\left(1-\frac{1}{n^2}\sin^2\theta\right)^{1/2}+\pi,\qquad \lambda=\frac{\lambda_0}{n} \]

  4. Exercise 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.
    NCERT’s answer
    (c)
    (c) Red and blue differ in frequency, so the two beams are incoherent. Their phase difference keeps changing, the cross term averages to zero, and the screen is evenly lit with no fringes.\[\Delta\phi(t)=(\omega_R-\omega_B)\,t+\phi_0 \]\[I=I_R+I_B+2\sqrt{I_RI_B}\,\langle\cos\Delta\phi\rangle=I_R+I_B,\qquad \langle\cos\Delta\phi\rangle=0 \]

  5. Exercise 5

    Figure 10.2\displaystyle 10.2 shows a standard two slit arrangement with slits S1, S2\displaystyle \mathrm{S}_1, \mathrm{~S}_2. P1,P2\displaystyle \mathrm{P}_1, \mathrm{P}_2 are the two minima points on either side of P (Fig. 10.2\displaystyle 10.2).
    At P2\displaystyle \mathrm{P}_2 on the screen, there is a hole and behind P2\displaystyle \mathrm{P}_2 is a second 2\displaystyle 2- slit arrangement with slits S3, S4\displaystyle \mathrm{S}_3, \mathrm{~S}_4 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.
    NCERT_Question_Class12_Physics_Exemplar_Ch10_Q5
    NCERT’s answer
    (d)
    (d) A finite-width hole passes light from either side of the minimum, so it acts as one coherent source for \(\displaystyle \mathrm{S_3}, \mathrm{S_4}\): regular two-slit fringes.Take \(\displaystyle d=\mathrm{S_1S_2}\), \(\displaystyle D\) the slits-to-screen distance, \(\displaystyle y\) measured from \(\displaystyle \mathrm{P}\) towards \(\displaystyle \mathrm{P_1}\).\[I(y)=4I_0\cos^2\frac{\delta}{2},\qquad \delta=\frac{2\pi d\,y}{\lambda D} \]\[\mathrm{S_1P_2}-\mathrm{S_2P_2}=\frac{d\,|y_2|}{D}=\frac{\lambda}{2}\ \ (\text{first minimum})\ \Rightarrow\ y_2=-\frac{\lambda D}{2d}\ \Rightarrow\ \delta=-\pi,\ \ I(y_2)=0 \]\[\frac{\delta}{2}=\frac{\pi d\,(y-y_2)}{\lambda D}-\frac{\pi}{2}\ \Rightarrow\ I(y)=4I_0\sin^2\!\left[\frac{\pi d\,(y-y_2)}{\lambda D}\right] \]\[I(y)>0\qquad \left(0<|y-y_2|<\frac{\lambda D}{d}\right) \]