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
View Answer
Diffraction is the bending of waves around obstacles or through apertures. Dispersion is the phenomenon where a wave's phase velocity depends on its frequency, leading to color separation. These are distinct phenomena. Both assertion and reason are 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
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 \( 45^\circ < \theta_B < 90^\circ \).