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
View Answer
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): Sound waves in air cannot be polarised.
Reason (R): Polarisation is the characteristic of light wave only.
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; sound waves in air are longitudinal waves and cannot be polarized. Reason (R) is false; polarization is a property of all transverse waves, not exclusively light waves.
Assertion (A): Two polaroids are crossed to each other. When either of them is rotated through \(30^\circ\), then only one eighth of the incident unpolarised light passes through the combination.
Reason (R): According to Malus’s law, \(I \propto cos^2 \theta\) where \(I\) is the resultant intensity transmitted and \(\theta\) is the angle between the optical axis of analyser and polariser.
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. Unpolarised light becomes \(I_0/2\) after the first polaroid. With \(60^\circ\) angle between axes, Malus's law gives \(I = (I_0/2)cos^2(60^\circ) = (I_0/2)(1/4) = I_0/8\). Reason (R) correctly states Malus's law, which explains (A).
Assertion (A): At the first glance the top surface of a Morpho’s butterfly’s wing appears a beautiful blue-green. If the wing moves, the colour changes.
Reason (R): Different pigments in the wing reflect light at different angles.
[Hint: It is due to interference of light rays reflected from different layers of wing.]
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. Morpho butterflies exhibit iridescence due to structural coloration. Reason (R) is false. The color is due to light interference by nanostructures on the wings, not pigments.
Assertion (A): Sound waves in air cannot be polarised.
Reason (R): Polarisation is the characteristic of light wave only.
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
Sound waves in air are longitudinal, meaning oscillations are parallel to propagation. Polarisation is a property of transverse waves where oscillations are perpendicular to propagation. Thus, sound cannot be polarised (A is true). Polarisation is characteristic of all transverse waves, not just light (R is false).
Assertion (A): Two polaroids are crossed to each other. When either of them is rotated through \(30^\circ\), then only one eighth of the incident unpolarised light passes through the combination.
Reason (R): According to Malus’s law, \(I \propto cos^2 \theta\) where \(I\) is the resultant intensity transmitted and \(theta\) is the angle between the optical axis of analyser and polariser.
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
When two crossed polaroids have one rotated by \(30^\circ\), the angle between their axes becomes \(60^\circ\). Incident unpolarised light \(I_0\) reduces to \(I_0/2\) after the first polaroid. By Malus's Law, \(I = (I_0/2) cos^2(60^\circ) = (I_0/2) (1/4) = I_0/8\). Both (A) and (R) are true, and (R) explains (A).
The Brewster’s angle \(\theta\) for an air-medium interface should be
1. 0° < \(\theta\) < 30°
2. 30° < \(\theta\) < 45°
3. 45° < \(\theta\) < 90°
4. \(\theta\) = 90°
View Answer
Brewster's law states \(tan\theta = \mu\). Since the refractive index of any medium with respect to air is \(\mu > 1\), we have \(tan\theta > 1 \implies \theta > 45^\circ\). Thus, \(45^\circ < \theta < 90^\circ\).
If light is entering from air to a medium and \( \theta_B \) represents Brewster’s angle for interface of the two medium, then
1. \( \theta_B = 90^\circ \)
2. \( 0^\circ < \theta_B < 90^\circ \)
3. \( 45^\circ < \theta_B < 90^\circ \)
4. \( 30^\circ < \theta_B < 60^\circ \)
View Answer
Brewster's law states that \( \tan\theta_B = \mu \). Since the light enters from air to a denser medium, \( \mu > 1 \), which means \( \tan\theta_B > 1 \) and hence \( 45^\circ < \theta_B < 90^\circ \).