Wave Optics - NEET Physics Questions
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Wave Optics

Question 51: easy

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.

Question 52: easy

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.

Question 53: easy

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.

Question 54: easy

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.

Question 55: easy

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.

Question 56: easy

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.

Question 57: easy

Assertion (A): If two sodium lamps are used illuminating two pinholes, interference fringes will not be observed.


Reason (R): Light waves coming from an ordinary source like sodium lamp are unpolarised in nature.


 

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: Two independent sources (like sodium lamps) are incoherent, meaning they do not maintain a constant phase relationship, thus cannot produce a stable interference pattern. Reason (R) is true: Light from ordinary sources like sodium lamps is unpolarised. However, incoherence is the primary reason for no interference, not the unpolarised nature. Thus, (R) is not the correct explanation for (A).

Question 58: 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.


 

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: 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.

Question 59: easy

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.


 

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: 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.

Question 60: easy

Assertion (A): In everyday life, we do not encounter diffraction of light in contrast to that for sound.


Reason (R): Diffraction characteristic is not exhibited by all kind of waves.


 

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: Observable diffraction occurs when wavelength is comparable to obstacle size. Light's wavelength is very small (nanometers), so its diffraction is not easily observed in daily life, unlike sound (wavelength in meters).


Reason (R) is false: Diffraction is a fundamental property of all waves, although its prominence depends on the wavelength and obstacle size. Hence, (A) is true, (R) is false.