Miscellaneous - NEET Physics Questions
Question 1: easy

Assertion (A): Resolving power of a microscope is different for different colours of illuminating light.


Reason (R): Resolving power of a microscope is directly proportional to the wavelength of illuminating 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

The resolving power (RP) of a microscope is given by \(RP = \frac{2NA}{1.22\lambda}\), where \(NA\) is the numerical aperture and \(\lambda\) is the wavelength. Since RP is inversely proportional to \(\lambda\), it will be different for different colours (different wavelengths). Reason (R) states direct proportionality, which is false.

Question 2: easy

Assertion (A): Distance between two coherent sources \(S_1\) and \(S_2\) is \(4\lambda\). A large circle is drawn around these sources with centre of circle lying on centre of \(S_1\) and \(S_2\). there are total 16 maxima on the circle.


Reason (R): Total number of minima on this circle are less compare to total number of maximas.

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

Given distance \(d=4lambda\). Path difference \(\Delta x = dcos\theta = 4\lambdacos\theta\). The range for \(\Delta x\) is \(-4\lambda \le \Delta x \le 4\lambda\). For maxima, \(\Delta x = n\lambda\), so \(-4 \le n \le 4\). This gives 9 integer values for \(n\). \(n=pm 4\) correspond to \(cos\theta = pm 1\) (2 points). The other 7 values \(n=pm 3, pm 2, pm 1, 0\) correspond to \(|costheta|<1\) (giving \(2times 7 = 14\) points). Total maxima = \(2+14=16\). So (A) is true. For minima, \(\Delta x = (n+1/2)\lambda\), so \(-4.5 \le n \le 3.5\). This gives 8 integer values for \(n\) (from -4 to 3). For all these values, \(|(n+1/2)/4|<1\), so each gives 2 points. Total minima = \(8\times 2 = 16\). Thus, there are 16 maxima and 16 minima. Reason (R) is false.

Question 3: easy

Assertion (A): If black strips of width \(w\) are separated by white strips of width \(d \left(d < w\right)\) are just distinguishable from a distance \(D\). Then resolving power of eye will be higher for smaller \(d\).


Reason (R): Resolution of eye is \(d/D\) and resolving power inversely proportional to resolution.

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 of the eye is \(RP = D/d\). Thus, for smaller \(d\), the resolving power \(RP\) will be higher. Reason (R) is false. While \(d/D\) represents the angular separation in this specific scenario when the strips are just distinguishable, it is not the general definition of the 'Resolution of eye'. Resolution of the eye is an intrinsic property (minimum angle it can resolve), not a variable specific to an observation.

Question 4: easy

Assertion (A): If light incident on surface of two different media. The refracted beam may be partially polarized.


Reason (R): If sum of angle of incidence and angle of refraction is \(\pi/2\) then a reflected light is totally polarised.


 

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. When unpolarized light is incident on the interface of two dielectric media, the refracted light is always partially polarized. Reason (R) is also true by Brewster's law; if the sum of the angle of incidence and refraction is \(90^circ\) (i.e., \(tan i_p = n\)), the reflected light is totally polarized. However, Reason (R) explains the total polarization of reflected light, not the partial polarization of refracted light, so R is not the correct explanation for A.

Question 5: easy

Assertion (A): If three polarisers are arranged such that the axis of any two successive polarisers make equal angle with each other. If unpolarised light of intensity \(I_0\) incident on first polariser then intensity of emergent light after 3rd polariser is \(\frac{I_0}{8}\). If angle between them is \(45^\circ\).


Reason (R): Each time intensity becomes \(50%\) by Malus law.


 

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. Intensity after 1st polariser is \(I_1 = I_0/2\). Given angle between successive polarisers is \(\theta = 45^\circ\). By Malus' Law, \(I_2 = I_1 \cos^2(45^\circ) = (I_0/2)(1/2) = I_0/4\). \(I_3 = I_2 \cos^2(45^\circ) = (I_0/4)(1/2) = I_0/8\). Reason (R) is false. Intensity becomes 50% only when \(cos^2\theta = 0.5\) (i.e., \(\theta = 45^\circ\)) and only for polarized light. The first polarizer reduces unpolarized light to 50% without \(cos^2\theta\) dependence. So, the general statement 'each time intensity becomes 50%' is false.

Question 6: easy

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. 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

The resolving power of a telescope is determined by the objective lens's diameter. The eyepiece's function is to magnify the image formed by the objective, not to enhance its resolution. Thus, details not resolved by the objective cannot be resolved by the eyepiece. Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Question 7: easy

Assertion (A): As angle subtended by the diameter of objective lens at the focus of microscope increased, resolving limit also increases.


Reason (R): Resolving limit proportional to tangent of the angle subtended by the diameter of objective lens at the focus of microscope.


 

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

The resolving limit of a microscope is \(\text{RL} = \frac{\lambda}{\text{2n} \sin\theta}\). As the angle \(\theta\) increases, \(sin\theta\) increases, causing \(\text{RL}\) to decrease (better resolution). So, Assertion (A) is false. Reason (R) is also false as \(\text{RL}\) is inversely proportional to \(sin\theta\), not proportional to \(tan\theta\).

Question 8: easy

Assertion (A): When refractive index of medium is increased resolving power also increases.


Reason (R): In medium of higher refractive index wavelength is higher.

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

The resolving power of a microscope is \(\text{R.P.} = \frac{\text{2n} \sin\theta}{\lambda}\) (wavelength in vacuum). It is directly proportional to refractive index \(\text{n}\), so (A) is true. Wavelength in a medium is \(\lambda_\text{medium} = \frac{\lambda_\text{vacuum}}{\text{n}}\). Higher \(\text{n}\)

Question 9: easy

Assertion (A): The resolving power of a telescope is more if the diameter of the objective in more.


Reason (R): Objective lens of larger focal length collect more light.


 

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

The resolving power of a telescope is \(\text{R.P.} = \frac{\text{D}}{\text{1.22}\lambda}\) where \(\text{D}\) is the diameter. Thus, larger \(\text{D}\) means higher \(\text{R.P.}), so (A) is true. Light collection depends on aperture (diameter), not directly on focal length. So, Reason (R) is false.

Question 10: easy

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).