Assertion (A): The stars which are not resolved in the image produced by the objective of a telescope can’t be further resolved by its eye piece.
Reason (R): The primary purpose of eyepiece of telescope is to provide the magnification of image produced by the objective.
1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer
Assertion (A) is true. The resolving power is determined by the objective; the eyepiece only magnifies the existing image, it cannot resolve features not already resolved by the objective. Reason (R) is true; the eyepiece's primary role is magnification. Reason (R) correctly explains Assertion (A).
Assertion (A): When a monochromatic light beam is incident normally on a reflective surface, under some condition it is possible that all lights is transmitted without any reflection.
Reason (R): When light after passing through a polaroid is incident on a reflecting surface at angle of incidence equals to polarizing angle, then all light gets transmitted without any reflection.
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 false. Total transmission at normal incidence on a reflective surface is only possible if the refractive indices are identical, implying no actual reflection.
Reason (R) is false. At Brewster's angle, only the p-polarized component of light is completely transmitted. If the light passed by the polaroid is s-polarized, it would be reflected. Therefore, the statement 'all light gets transmitted' is not universally true for light passed by a polaroid without specifying its polarization.
Thus, both (A) and (R) are false.
Assertion (A): Wave nature can be proved by phenomena of interference and diffraction.
Reason (R): Only transverse wave can show the phenomena of polarization.
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. Interference and diffraction are characteristic wave phenomena, providing strong evidence for the wave nature of light.
Reason (R) is true. Polarization is a property exclusive to transverse waves, where oscillations are perpendicular to the propagation direction.
Reason (R) describes a unique property of transverse waves, which is distinct from demonstrating wave nature via interference/diffraction. Thus, (R) does not explain (A).
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): 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): 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).