Practice NEET Young's Double Slit Experiment Questions
Question 51:
easy
Assertion (A): The central fringe is bright or dark, it dependents on the initial phase difference between the two coherent sources.
Reason (R): The pattern and position of fringes always remains same even after the introduction of transparent medium in a path of one of the slit.
Assertion (A) is true. The nature of the central fringe depends on the initial phase difference. Reason (R) is false. Introducing a transparent medium causes a path difference \((n-1)t\) and shifts the entire fringe pattern.
Assertion (A): Diffraction is a sure indication of wave nature.
Reason (R): Only transverse waves can be diffracted.
Assertion (A) is true; diffraction is a characteristic property of all waves. Reason (R) is false; both transverse (e.g., light) and longitudinal (e.g., sound) waves can be diffracted.
Assertion (A): The best contrast of the interference pattern is obtained when the intensity of the emerging lights from the two slits of the Young’s experimental set-up are equal.
Reason (R): Intensity is proportional to the square of the amplitude.
For best contrast in an interference pattern, the amplitudes of the interfering waves must be equal, implying equal intensities. Intensity \(I\) is proportional to the square of the amplitude \(A\), i.e., \(I \propto A^2\). Thus, equal intensities lead to maximum contrast.
Assertion (A): The central fringe is bright or dark, it depends on the initial phase difference between the two coherent sources.
Reason (R): The pattern and position of fringes always remains same even after the introduction of transparent medium in a path of one of the slit.
The central fringe's nature (bright/dark) depends on the initial phase difference. Introducing a transparent medium with refractive index \(n\) and thickness \(t\) in one path creates an additional path difference \((n-1)t\), shifting the entire fringe pattern. So, (R) is false.
The intensity at the maximum in a Young’s double slit experiment is $I_0$. Distance between two slits is $d = 5\lambda$, where $lambda$ is the wavelength of light used in the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance $D = 10d$?
(2016-I)
The point in front of one slit has $y = d/2$. The path difference is $Delta x = frac{yd}{D} = frac{(d/2)d}{10d} = frac{d}{20}$. Since $d = 5lambda$, $Delta x = frac{5lambda}{20} = frac{lambda}{4}$. The phase difference is $Deltaphi = frac{2pi}{lambda} Delta x = frac{pi}{2}$. The intensity is $I = I_0 cos^2(frac{Deltaphi}{2}) = I_0 cos^2(frac{pi}{4}) = frac{I_0}{2}$.
The interference pattern is obtained with two coherent light sources of intensity ratio n. In the interference pattern, the ratio $\frac{I_{max} – I_{min}}{I_{max} + I_{min}}$ will be:
(2016 – II)
The maximum and minimum intensities are $I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2$ and $I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2$. Given $I_1/I_2 = n$, we can write $I_1 = n I_2$. The required ratio is $\frac{(\sqrt{n}+1)^2 - (\sqrt{n}-1)^2}{(\sqrt{n}+1)^2 + (\sqrt{n}-1)^2}$. Expanding the squares gives $\frac{4\sqrt{n}}{2(n+1)} = \frac{2\sqrt{n}}{n+1}$.
Two periodic waves of intensities $I_1$ and $I_2$ pass through a region at the same time in the same direction. The sum of the maximum and minimum intensities is:
(2008)
The maximum intensity is $I_{max} = I_1 + I_2 + 2\sqrt{I_1 I_2}$ and the minimum intensity is $I_{min} = I_1 + I_2 - 2\sqrt{I_1 I_2}$. Their sum is $I_{max} + I_{min} = 2(I_1 + I_2)$.
Ratio of intensities of two waves are given by 4 : 1. Then ratio of the amplitudes of the two waves is:
(1991)
The intensity of a wave is directly proportional to the square of its amplitude ($I \propto A^2$). Therefore, the ratio of amplitudes is $A_1 / A_2 = \sqrt{I_1 / I_2} = \sqrt{4 / 1} = 2 / 1$.
In a Young’s double slit experiment, a student observes 8 fringes in a certain segment of screen when a monochromatic light of 600 nm wavelength is used. If the wavelength of light is changed to 400 nm, then the number of fringes he would observe in the same region of the screen is
(2022)
The width of the screen segment remains constant. Thus, $n_1 \beta_1 = n_2 \beta_2$, which implies $n_1 \lambda_1 = n_2 \lambda_2$. Substituting the values: $8 \times 600 = n_2 \times 400$, giving $n_2 = 12$ fringes.
In Young’s double slit experiment, if the separation between coherent sources is halved and the distance of the screen from the coherent sources is doubled, then the fringe width becomes:
(2020)
Fringe width is given by $beta = frac{lambda D}{d}$. When $D' = 2D$ and $d' = d/2$, the new fringe width is $beta' = frac{lambda (2D)}{(d/2)} = 4 frac{lambda D}{d} = 4beta$. It becomes four times the original.