Q.A beam of light consisting of two wavelengths, $650\ \text{nm}$ and $520\ \text{nm}$, is used to obtain interference fringes in a Young's double-slit experiment.

Showing the 12 most recent of 13 on this concept.

1. GUJCET 2026Set x1 markMCQ
   Q.In a Young's double slit experiment, the slits are separated by 0.2 mm and placed 2.0 m away. The distance between the central bright fringe and the first bright fringe (fringe width) is measured to be 1.5 cm. Determine the wavelength of light used in the experiment. (A) 4200 $\text{\AA}$ (B) 5000 $\text{\AA}$ (C) 4600 $\text{\AA}$ (D) 15000 $\text{\AA}$ (Calculated result)

   ›Reveal solutionSolution

   [!TLDR] Substituting the given numbers into $\lambda=\beta d/D$ yields $1.5\times10^{-6}$ m $=15000\,\text{\AA}$, matching option (D).

   Concept

   In Young's double-slit experiment the fringe width is $\beta = \dfrac{\lambda D}{d}$, where $d$ is the slit separation and $D$ the slit-to-screen distance. Rearranged, $\lambda = \dfrac{\beta d}{D}$.

   Solution

   Given $d = 0.2\,\text{mm} = 0.2\times10^{-3}$ m, $D = 2.0$ m, $\beta = 1.5\,\text{cm} = 1.5\times10^{-2}$ m. …

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2. GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQ
   Q.If path difference between two waves is 3 lambda then phase difference related with them is ___.

   (a) 4 pi rad

   (b) 3 pi rad

   (c) 2 pi rad

   (d) 6 pi rad

   ›Reveal solutionSolution

   Path difference and phase difference are related by delta(phi) = (2 pi/lambda) x (path difference), because one full wavelength of path corresponds to a full 2 pi cycle of phase.

   …

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3. GUJCET 2025Set 031 markMCQ
   Q.Two waves of same intensity $I_{0}$ emitted from two sources having same phase difference ($\phi$). Due to superposition of two waves, the intensity of resultant wave is directly proportional to ______. (A) $\sin^{2}\left(\frac{\phi}{2}\right)$ (B) $\sin^{2}\phi$ (C) $\cos^{2}\left(\frac{\phi}{2}\right)$ (D) $\cos^{2}\phi$

   ›Reveal solutionSolution

   Superposition of two equal-intensity coherent waves gives $I=4I_0\cos^2(\phi/2)$.

   Concept — interference. Resultant intensity of two waves of intensities $I_1,I_2$ with phase difference $\phi$ is $I=I_1+I_2+2\sqrt{I_1I_2}\cos\phi$.

   Steps.

   • With $I_1=I_2=I_0$: $I=2I_0+2I_0\cos\phi=2I_0(1+\cos\phi)$. …

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4. GUJCET 2025Set 031 markMCQ
   Q.In Young's double slit experiment, the slits are separated by 0.54 mm and the screen is placed 1.8 m away. The distance between central bright fringe and sixth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment. (A) 5000 Å (B) 600 nm (C) 8000 nm (D) 800 nm

   ›Reveal solutionSolution

   [!TLDR]

   From $\lambda = x_n d/(nD)$ the wavelength is $600\,\text{nm}$.

   Concept

   In Young's double-slit experiment the distance of the $n$-th bright fringe from the centre is $x_n = \dfrac{n\lambda D}{d}$, so $\lambda = \dfrac{x_n\,d}{n\,D}$.

   Solution

   Given $d = 0.54\,\text{mm} = 0.54\times10^{-3}\,\text{m}$, $D = 1.8\,\text{m}$, $x_6 = 1.2\,\text{cm} = 1.2\times10^{-2}\,\text{m}$, $n = 6$: …

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5. GUJCET 2024Set 131 markMCQ
   Q.In a Young's double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Then the wavelength of light used in the experiment is ________. (A) 500 nm (B) 660 nm (C) 600 nm (D) 550 nm

   ›Reveal solutionSolution

   Fourth bright fringe: $y_4 = \frac{4\lambda D}{d}$, solve for $\lambda$.

   Steps. $d=0.28\times10^{-3}$ m, $D=1.4$ m, $y_4 = 1.2\times10^{-2}$ m, $n=4$. …

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6. GUJCET 2023Set 091 markMCQ
   Q.Two slits are made 10 millimeter apart and the screen is placed 1.5 metre away. What is the fringe separation when a wavelength of $7000\,\text{\AA}$ is used? (A) $105\,\mu m$ (B) $1.05\,\mu m$ (C) $10.5\,\mu m$ (D) $0.105\,\mu m$

   ›Reveal solutionSolution

   [!TLDR]

   Using $\beta = \lambda D/d$ gives a fringe width of $105\,\mu m$.

   Concept

   In Young's double-slit experiment the spacing between adjacent bright (or dark) fringes is $\beta = \dfrac{\lambda D}{d}$, where $D$ is the slit-to-screen distance and $d$ the slit separation.

   Solution

   Convert units: $\lambda = 7000\,\text{\AA} = 7\times10^{-7}\,\text{m}$, $D = 1.5\,\text{m}$, $d = 10\,\text{mm} = 1\times10^{-2}\,\text{m}$. …

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7. GSEB Higher Secondary Certificate (HSC) Examination 2023Set ANNUAL1 markMCQ
   Q.In Young's double experiment the distance between two slits is 0.2 mm and the distance between slit and screen is 1.5 m. The wavelength of light used is 600 nm. The distance between any two consecutive bright fringes is ___ mm.

   (a) 0.8

   (b) 4.5

   (c) 0.5

   (d) 2.0

   ›Reveal solutionSolution

   The spacing of consecutive bright fringes is beta = lambda D/d = 4.5 mm.

   Fringe width in Young's double-slit experiment:

   beta = lambda D / d.

   Substitute lambda = 600 nm = 600 x 10^-9 m, D = 1.5 m, d = 0.2 mm = 0.2 x 10^-3 m: …

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8. GUJCET 2022Set 171 markMCQ
   Q.Two slits are made 3 millimetre (3 mm) apart and the screen is placed 2 m away. What is the fringe separation when blue-green light of wavelength 600 nm is used? (A) 0.4 mm (B) 0.6 mm (C) 0.5 mm (D) 0.7 mm

   ›Reveal solutionSolution

   $\beta = \dfrac{\lambda D}{d}$.

   Steps.

   • $\lambda = 600\text{ nm} = 6\times10^{-7}$ m, $D = 2$ m, $d = 3\text{ mm} = 3\times10^{-3}$ m. …

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9. GUJCET 2021Set 151 markMCQ
   Q.The wavelength of light 500 nm is used in a Young's double-slit experiment. The distance between the slits and screen is 100 cm and the slits are separated by 1 mm. Then find distance between fifth (5th) and third (3rd) bright fringes. (A) 1 mm (B) 3 mm (C) 2 mm (D) 4 mm

   ›Reveal solutionSolution

   Consecutive bright fringes are beta = lambda D/d apart; the 5th and 3rd differ by 2 beta.

   Concept. $\beta=\dfrac{\lambda D}{d}$, and $x_5-x_3=(5-3)\beta=2\beta$. …

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10. GUJCET 2020Set 071 markMCQ
    Q.The distance between two slits is 3 mm & screen is placed at 2 m distance. When blue-green light of wavelength 500 nm is used then distance between two fringes will be? (A) 0.5 mm (B) 0.43 mm (C) 0.33 mm (D) 0.4 mm

    ›Reveal solutionSolution

    Fringe spacing in YDSE is $\beta=\dfrac{\lambda D}{d}$.

    Concept: In Young's double-slit experiment the distance between adjacent bright (or dark) fringes is $\beta=\dfrac{\lambda D}{d}$. …

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11. GUJCET 2019Set 131 markMCQ
    Q.In Young's experiment fourth bright fringe produced by light of $5000\,\text{Å}$ superposes on the fifth bright fringe of an unknown wavelength. the unknown wavelength is ............ $\text{Å}$ . (A) 8000 (B) 5000 (C) 6000 (D) 4000

    ›Reveal solutionSolution

    Overlapping bright fringes have equal positions, so $n_1\lambda_1 = n_2\lambda_2$.

    Concept first: The position of the $n$-th bright fringe is $y_n = n\lambda D/d$. If the 4th bright fringe of $5000\,\text{Å}$ coincides with the 5th bright fringe of the unknown wavelength: …

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12. GUJCET 2015Set C1 markMCQ
    Q.Light of wave length $\lambda$ is incident on slit of width d. The resulting diffraction pattern is observed on a screen placed at distance D. The linear width of central maximum is equal to width of the slit, then D = _____ (A) $\dfrac{2\lambda^2}{d}$ (B) $\dfrac{d^2}{2\lambda}$ (C) $\dfrac{d}{\lambda}$ (D) $\dfrac{2\lambda}{d}$

    ›Reveal solutionSolution

    [!TLDR] Setting the central-maximum width equal to the slit width gives $D = \dfrac{d^2}{2\lambda}$, option (B).

    Concept …

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