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Wave Optics
Young's Double Slit Experiment
8 questions from 8 papers, each with its answer.
- questions
- 8
- papers
- 8
- years asked, 2017–2026
- 5 of 10
- of Wave Optics
- 20%
Appears in
Concepts tested
- angular fringe separation
- angular fringe width in a medium
- Coincidence of bright fringes
- fringe number vs wavelength
- fringe width dependence on d and D
- Intensity and phase difference
- Intensity ratio
- YDSE with white light source
All Questions
2024 1 question
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If the monochromatic source in Young's double slit experiment is replaced by white light, thenWhite light is a mixture of different wavelengths (colours). At the centre, the path difference is zero for all wavelengths, so all colours combine to give a central bright white fringe. Away from the centre, different colours have their maxima at different positions, so the surrounding fringes appear coloured, and only a few coloured fringes are visible.
2023 1 question
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For Young's double slit experiment, two statements are given below: Statement I: If screen is moved away from the plane of slits, angular separation of the fringes remains constant. Statement II: If the monochromatic source is replaced by another monochromatic source of larger wavelength, the angular separation of fringes decreases. In the light of the above statements, choose the correct answer from the options given below:
$\Delta\theta=\frac{\lambda}{d}$ depends on $\lambda$ but not on the screen distance.
$\Delta\theta=\frac{\lambda}{d}$ $\Delta\theta$ is proportional to $\lambda$ but independent of $D$.
2022 1 question
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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 isFringe position $y = n\lambda\frac{D}{d}$; for the same segment of the screen, $n_1\lambda_1 = n_2\lambda_2$ $8 \times 600 = n_2 \times 400$ $n_2 = 12$
2020 1 question
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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:Fringe width $\beta = \frac{\lambda D}{d}$ $d' = \frac{d}{2}$ and $D' = 2D$ ⇒ $\beta' = \frac{\lambda(2D)}{d/2} = \frac{4\lambda D}{d} = 4\beta$ So the fringe width becomes four times.
2019 1 question
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In a double slit experiment, when light of wavelength 400 nm was used, the angular width of the first minima formed on a screen placed 1 m away, was found to be $0.2^\circ$. What will be the angular width of the first minima, if the entire experimental apparatus is immersed in water ($\mu_{water}=4/3$)?In air, angular fringe width $\theta_0=\frac{\lambda}{d}$. In water, the wavelength becomes $\frac{\lambda}{\mu}$, so $\theta_w=\frac{\theta_0}{\mu}$ $\theta_w=\frac{0.2^\circ}{4/3}=0.15^\circ$
2015 AIPMT-II 1 question
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Two slits in Young's experiment have widths in the ratio 1 : 25. The ratio of intensity at the maxima and minima in the interference pattern, $\frac{I_{max}}{I_{min}}$ is:Intensity ∝ slit width: $\frac{I_1}{I_2} = \frac{1}{25} \Rightarrow \sqrt{\frac{I_2}{I_1}} = 5$
$\frac{I_{max}}{I_{min}} = \left(\frac{\sqrt{I_2} + \sqrt{I_1}}{\sqrt{I_2} - \sqrt{I_1}}\right)^2 = \left(\frac{5 + 1}{5 - 1}\right)^2 = \frac{9}{4}$
2014 AIPMT 1 question
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In the Young's double-slit experiment, the intensity of light at a point on the screen where the path difference is $\lambda$ is K, ($\lambda$ being the wavelength of light used). The intensity at a point where the path difference is $\frac{\lambda}{4}$, will be:Path difference $\lambda$ → phase difference $2\pi$; path difference $\frac{\lambda}{4}$ → phase difference $\frac{\pi}{2}$.
$K = 4I_0$
$K' = 4I_0\cos^2\left(\frac{\pi}{4}\right) = 2I_0 = \frac{K}{2}$
2013 NEET 1 question
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In Young's double slit experiment, the slits are 2 mm apart and are illuminated by photons of two wavelengths $\lambda_1$ = 12000 Å and $\lambda_2$ = 10000 Å. At what minimum distance from the common central bright fringe on the screen 2 m from the slit will a bright fringe from one interference pattern coincide with a bright fringe from the other?For coincidence, $n_1\lambda_1 = n_2\lambda_2$
$\frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1} = \frac{10000}{12000} = \frac{5}{6}$, so the minimum $n_1$ and $n_2$ are 5 and 6.
$x_{min} = \frac{n_1\lambda_1D}{d} = \frac{5(12000\times10^{-10})(2)}{2\times10^{-3}} = 6\times10^{-3}$ m = 6 mm