Assertion (A): If incident wavefront is plane, then after reflection or refraction the emerging wave front also must be plane.
Reason (R): Wavefronts are in the direction of energy propagation by light.
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: A plane wavefront remains plane after reflection or refraction from plane/spherical surfaces. Reason (R) is false: Wavefronts are surfaces of constant phase and are perpendicular to the direction of energy propagation (rays). Hence, (A) is true, but (R) is false.
Assertion (A): Light is a wave phenomenon.
Reason (R): Light requires a material medium for its propagation.
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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Light is an electromagnetic wave and can travel through vacuum, so it does not require a material medium for propagation. Thus, Assertion (A) is true but Reason (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.
Assertion (A): In Young’s double slit experiment, if one of the slits is closed, intensity at the position of central fringe becomes half.
Reason (R): Resultant intensity, being sum of intensities from individual slits, becomes half as one slit is closed.
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 intensity from one slit is (I_0), the central maximum with two slits is (4I_0). If one slit is closed, the intensity at the center becomes (I_0), which is one-fourth, not half. Reason (R) is also incorrect as intensities don't simply add algebraically. Both A and R are false.
Assertion (A): In YDSE, fringes with blue light are thicker than those for red light.
Reason (R): In YDSE, the \(n^{\text{th}}\) maxima always comes before the \(n^{\text{th}}\) minima.
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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Fringe width \(\beta = \frac{\lambda D}{d}\). Since \(\lambda_{\text{blue}} < \lambda_{\text{red}}\), blue fringes are thinner than red fringes, so A is false. Minima and maxima alternate, and the \(n^{\text{th}}\) dark fringe typically occurs closer to the central maximum than the \(n^{\text{th}}\) bright fringe (for (n ge 1)). So R is also false.
Assertion (A): We cannot get diffraction pattern from a wide slit illuminated by monochromatic light.
Reason (R): In diffraction pattern, all the bright bands are not of the same intensity.
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 in a practical sense; a very wide slit (\(a \gg lambda\)) yields a pattern with very small angular spread, making it indiscernible. Reason (R) is true; the intensity of secondary maxima decreases rapidly. (R) does not explain (A).
Assertion (A): Diffraction of light is due to dispersion.
Reason (R): Change in path of light around “the corners separates the wavelength of various colours.
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 false; diffraction is the bending of waves, while dispersion is wavelength-dependent refractive index. Reason (R) is false; color separation in diffraction is due to \(\theta \propto \lambda\), not dispersion around corners.