Practice NEET Young's Double Slit Experiment Questions
Question 41:
easy
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.
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.
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 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.
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.
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): 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.
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.
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): 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.
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.
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.
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): The best contrast of the interference pattern is obtained when the intensity of the emerging lights from the two slits of the Young’s experimental set-up are equal.
Reason (R): Intensity is proportional to the square of the amplitude.
Assertion (A) is true, as equal intensities (\(I_1 = I_2\)) ensure \(I_{text{min}} = 0\) for maximum contrast.
Reason (R) is true, as intensity \(I propto A^2\). However, (R) does not explain why \(I_1 = I_2\) leads to best contrast.