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.
Assertion (A): When a width of one of the slits of Young’s double slit experiment is double that of the other than brighter fringes are nine times brighter than the dark fringes.
Reason (R): The amplitude of the wave is proportional to the width of the slit.
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
In Young's double slit experiment, intensity is proportional to the square of amplitude \(I \propto A^2\). The amplitude \(A\) is proportional to the slit width \(w\). If \(w_1 = 2w_2\), then \(A_1 = 2A_2\). The ratio of maximum to minimum intensity is \(I_{max}/I_{min} = ((A_1+A_2)/(A_1-A_2))^2\). Substituting \(A_1 = 2A_2\), we get \(I_{max}/I_{min} = ((2A_2+A_2)/(2A_2-A_2))^2 = (3A_2/A_2)^2 = 3^2 = 9\). So, (A) is true. (R) is also true, as amplitude is proportional to slit width. And (R) correctly explains (A).
Assertion (A): When tiny circular obstacle is placed in the path of light from some distance, a bright spot is seen at the centre of shadow of the obstacle.
Reason (R): Constructive interference occurs at the centre of the shadow.
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) describes the Poisson's spot (or Arago spot) phenomenon, a classical example of diffraction where a bright spot appears in the center of the shadow of an opaque circular object. Reason (R) correctly states that this occurs due to constructive interference of light waves diffracting around the edges of the obstacle and meeting in phase at the center of the shadow. Both are true and R explains A.
Assertion (A): If width of one of the slit in YDSE is slightly increased, then maximum and minimum both Intensity will increase.
Reason (R): Intensity reaching from that slit on screen will slightly increase.
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
In YDSE, the intensity of light from a slit is proportional to its width \(I \propto w\). Also, amplitude \(A \propto \sqrt{I}\). If one slit's width increases, its amplitude \(A\) increases. The maximum intensity is \(I_{max} = (A_1+A_2)^2\) and minimum intensity is \(I_{min} = (A_1-A_2)^2\). If \(A_1\) increases, both \(I_{max}\) and \(I_{min}\) will increase. Both A and R are true, and R explains A.
Assertion (A): If white light is used in place of monochromatic light in YDSE then central point is white. Although at other places coloured fringes will be obtained.
Reason (R): At centre path difference is zero for all wave lengths. Hence all wave will interfere constructively.
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
In Young's double slit experiment, the central point has a path difference of zero \(\Delta x = 0\) for all wavelengths \(\lambda\). The condition for constructive interference is \(\Delta x = n\lambda\). For \(n=0\), \(\Delta x = 0\) which is true for all \(\lambda\). Therefore, all colors interfere constructively at the center, resulting in a white fringe. Other fringes are colored due to dispersion. Both are true and R explains A.
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.
Assertion (A): As the separation between the two slits is increased width of fringes decreases.
Reason (R): On increasing separation between two slits, angular separation of fringes decreases.
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
Fringe width \(beta = \frac{\lambda D}{d}\). As slit separation \(d\) increases, fringe width \(beta\) decreases. Angular fringe width \(theta = \frac{\lambda}{d}\). As \(d\) increases, angular separation \(\theta\) decreases. Thus, R is the correct explanation of A.
Assertion (A): In case of Young double slit experiment width of all fringes were equal.
Reason (R): Angular width of fringes were equal.
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
In Young's Double Slit Experiment (YDSE), the fringe width \(beta = \frac{\lambda D}{d}\) is constant for all fringes, hence they are of equal width. The angular fringe width \(\theta = \frac{\lambda}{d}\) is also constant. Reason (R) correctly explains Assertion (A).
Assertion (A): In case of single slit diffraction intensity of higher order maxima decreases.
Reason (R): Higher order maxima are at larger distance.
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 intensity of higher order maxima in single slit diffraction decreases rapidly. Reason (R) is also true; higher order maxima occur at larger distances from the central maximum. However, R is not the correct explanation for A, as the decrease in intensity is due to the smaller effective aperture contributing to these maxima, not their distance.
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.