Assertion (A): Diffraction takes place for all types of waves mechanical or non-mechanical, transverse or longitudinal.
Reason (R): Diffraction’s effects are perceptible only if wavelength of wave is comparable to dimensions of diffracting device.
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
View Answer
Both Assertion (A) and Reason (R) are true. Diffraction is a universal wave phenomenon, occurring for all wave types. Its effects are most noticeable when the wavelength is comparable to the obstacle's size.
However, (R) states the condition for observation, not the fundamental reason why diffraction occurs for all waves (A).
Assertion (A): Light is diffracted around the edges of obstacles and it bend such a way which is not easily observed.
Reason (R): The wavelength of light is very small.
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
View Answer
Diffraction is noticeable when the wavelength is comparable to the obstacle size. Light has a very small wavelength \( approx 400-700 \text{ nm}\), making its diffraction around macroscopic objects hard to observe. Hence, both A and R are true, and R is the correct explanation for A.
Assertion (A): In Young’s double slit experiment if intensity of each source is \(I_0\) then minimum and maximum intensity is zero and \(4I_0\) respectively.
Reason (R): In Young’s double slit experiment energy conservation is not followed.
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
View Answer
In YDSE with coherent sources of intensity (I_0) each, \(I_{\text{min}} = (\sqrt{I_0} - \sqrt{I_0})^2 = 0\) and \(I_{\text{max}} = (\sqrt{I_0} + \sqrt{I_0})^2 = 4I_0\). Thus, A is true. Energy is conserved in interference; it's redistributed, not destroyed. Thus, R is false.
Assertion (A): Radio waves cannot be diffracted by the buildings.
Reason (R): The wavelength of radio waves is very small.
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
View Answer
Radio waves have wavelengths comparable to or larger than buildings \( \text{meters to kilometers}\), enabling them to diffract easily around obstacles. Thus, A is false. The wavelength of radio waves is large, not small. Thus, R is false.
Assertion (A): In standard YDSE set up with visible light, the position on screen where phase difference is zero appears bright.
Reason (R): In YDSE set up amplitude of electromagnetic field at central bright fringe is not varying with time.
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
View Answer
A zero phase difference signifies constructive interference, resulting in a bright fringe. Hence, A is true. The amplitude of the electromagnetic field at the central bright fringe remains constant over time in a stable interference pattern, but this is not the reason for it being bright due to zero phase difference. Hence, R is true but not the correct explanation for A.
Assertion (A): In Young’s experiment, the fringe width for dark fringes is different from that for bright fringes.
Reason (R): In Young’s double slit experiment with a source of white light, only black and white fringes are observed.
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
View Answer
In YDSE, the fringe width is given by \(\beta = \frac{\lambda D}{d}\), which is independent of whether the fringe is bright or dark. Hence, A is false. With white light, a central bright white fringe is formed, and then colored fringes are observed, not just black and white. Hence, R is false.
Assertion (A): The plane of polarization of reflected ray is parallel to the refracting surface, when light is incident at polarising angle.
Reason (R): Vibration of electric field in refracted ray ceases about plane parallel to refracting surface.
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
View Answer
At the polarizing angle (Brewster's angle), the reflected light is completely plane-polarized with its electric field vibrations perpendicular to the plane of incidence (i.e., parallel to the refracting surface). Thus, A is true. The refracted ray is partially polarized and still has electric field vibrations in various planes, not ceasing in any specific plane. Thus, R is false.
Assertion (A): Diffraction is common in sound but not common in light waves.
Reason (R): Wavelength of light wave is more than the wavelength of sound.
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
View Answer
Diffraction is pronounced when the wavelength is comparable to the obstacle size. Sound waves have large wavelengths (\( \text{meters}\)), making their diffraction common around everyday objects. Light waves have very small wavelengths (\( \text{nanometers}\)), so their diffraction is less observed. Thus, A is true. The wavelength of light is significantly smaller than the wavelength of sound. Thus, R is false.
Assertion (A): If a glass slab is placed in front of one of the slits, then fringe width will decrease.
Reason (R): Glass slab will produce no path difference.
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
View Answer
Placing a glass slab in front of one slit causes a path difference of \(t(\mu - 1)\) and shifts the entire fringe pattern but does not alter the fringe width \(\beta = \frac{\lambda D}{d}\). Thus, A is false. A glass slab indeed introduces an additional optical path difference. Thus, R is false.
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
View Answer
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.