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): 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.
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.
Assertion (A): On increasing wavelength of light used, resolving power increases.
Reason (R): On increasing wavelength, width of central maxima 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
Assertion (A) is false. Resolving power of optical instruments (e.g., telescope, microscope) is inversely proportional to the wavelength (\(RP \propto 1/\lambda\)). So, increasing wavelength decreases resolving power. Reason (R) is also false. In diffraction, the width of the central maximum is directly proportional to the wavelength (\(w \propto \lambda\)). Thus, increasing wavelength increases the width of the central maximum.
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.
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).