Assertion (A): If two sodium lamps are used illuminating two pinholes, interference fringes will not be observed.
Reason (R): Light waves coming from an ordinary source like sodium lamp are unpolarised in nature.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
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Assertion (A) is true: Two independent sources (like sodium lamps) are incoherent, meaning they do not maintain a constant phase relationship, thus cannot produce a stable interference pattern. Reason (R) is true: Light from ordinary sources like sodium lamps is unpolarised. However, incoherence is the primary reason for no interference, not the unpolarised nature. Thus, (R) is not the correct explanation for (A).
Assertion (A): In a Young’s double slit experiment (YDSE), if the screen is move away from the plane of slits, Angular fringe width remains unchanged.
Reason (R): Linear and Angular fringe width is directly proportional to D.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
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Assertion (A) is true: Angular fringe width is given by \(\theta = \frac{\lambda}{d}\), which is independent of \(D\) (distance to screen). Reason (R) is false: While linear fringe width \(\beta = \frac{\lambda D}{d}\) is proportional to \(D\), angular fringe width \(\theta\) is not. Hence, (A) is true, (R) is false.
Assertion (A): In case of YDSE, if monochromatic light is replaced by white light then closest on either side of central white fringe will be blue and farthest will appear red.
Reason (R): Fringe width for blue will be greater than that for red for same bright fringe.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
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Assertion (A) is true: Fringe width is \(\beta = \frac{\lambda D}{d}\). Since \(\lambda_{\text{red}} > \lambda_{\text{blue}}\), it follows that \(\beta_{\text{red}} > \beta_{\text{blue}}\). Thus, red fringes are wider and appear farther from the center, while blue fringes are closer. Reason (R) is false: Fringe width for blue light is smaller than that for red light. Hence, (A) is true, (R) is false.
Assertion (A): In everyday life, we do not encounter diffraction of light in contrast to that for sound.
Reason (R): Diffraction characteristic is not exhibited by all kind of waves.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
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Assertion (A) is true: Observable diffraction occurs when wavelength is comparable to obstacle size. Light's wavelength is very small (nanometers), so its diffraction is not easily observed in daily life, unlike sound (wavelength in meters).
Reason (R) is false: Diffraction is a fundamental property of all waves, although its prominence depends on the wavelength and obstacle size. Hence, (A) is true, (R) is false.
Assertion (A): In the double slit experiment, if one of the slit is closed, no fringe pattern will be observed on the screen.
Reason (R): In interference, phenomenon of diffraction is also included.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
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Assertion (A) is true: If one slit is closed, the two-source condition for interference is not met, so a distinct interference fringe pattern is not observed. Instead, a single-slit diffraction pattern appears. Reason (R) is true: The observed double-slit intensity pattern is a combination of interference from two slits and diffraction from each individual slit. Both (A) and (R) are true, but (R) is not the correct explanation for (A).
Assertion (A): Incoherent sources do not produce an interference pattern.
Reason (R): Light from two coherent sources that are not in phase does not produce an interference pattern.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
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Assertion (A) is true: Incoherent sources have rapidly fluctuating phase differences, resulting in an average uniform intensity rather than a stable interference pattern.
Reason (R) is false: Coherent sources, even if not in phase (i.e., having a constant non-zero phase difference), will still produce a stable interference pattern, though its position might shift. Hence, (A) is true, (R) is false.
Assertion (A): Two sources of light emit light waves of same frequency but of different amplitudes. Also the phase difference between light waves from the two sources at any point is time independent. Therefore, observable interference will be obtained when light waves from the two sources superimpose.
Reason (R): The sources are not coherent due to unequal amplitudes.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
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For observable interference, sources must have a constant phase difference (coherent) and same frequency. Different amplitudes still allow interference, just with non-zero minimum intensity. Coherence is related to phase difference, not amplitude equality. Thus, A is true and R is false.
Assertion (A): Interference pattern is obtained on a screen due to two identical coherent sources of monochromatic light. The intensity at the central part of the screen becomes one-fourth if one of the sources is blocked.
Reason (R): The resultant intensity at any point is the algebraic sum of the intensities due to two sources.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
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For two identical coherent sources of intensity \(I_0\) each, the central maximum intensity is \(4I_0\). If one source is blocked, the intensity becomes \(I_0\), which is one-fourth of \(4I_0\). The resultant intensity in interference is not an algebraic sum of individual intensities but depends on the phase difference. Thus, A is true and R is false.
Assertion (A): In Young’s double slit experiment, assuming slits to be of equal widths, intensity at interference maxima is four times the intensity due to each slit.
Reason (R): Intensity is proportional to the square of amplitude.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
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If \(I_0\) is the intensity from each slit, then the amplitude is \(A_0 \propto \sqrt{I_0}\). At maxima, amplitudes add to \(2A_0\), so intensity is \((2A_0)^2 \propto 4A_0^2 = 4I_0\). Intensity is indeed proportional to the square of amplitude, explaining this result. Both A and R are true, and R explains A.
Assertion (A): If Young’s double slit experiment is performed with white light, the bright fringes produced are white and the dark fringes black.
Reason (R): In case of interference, there is no colour splitting.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
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When white light is used in YDSE, the central fringe is white. However, other bright fringes are coloured due to dispersion (different wavelengths have different fringe widths). Dark fringes are also not perfectly black. Thus, there is colour splitting. Both A and R are false.