Q.In a double-slit experiment the angular width of a fringe is found to be $0.2^\circ$ on a screen placed $1\ \text{m}$ away. The wavelength of light used is $600\ \text{nm}$. What will be the angular width of the fringe if the entire experimental apparatus is immersed in water? Take refractive index of water to be $\frac{4}{3}$.

Showing the 12 most recent of 43 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.Two waves of amplitude 'A' and frequency 'ν' are superimposed with each other. After super position, the maximum intensity is (A) A (B) $A^2$ (C) 2A (D) $4A^2$

   ›Reveal solutionSolution

   This tests superposition of waves: constructive interference gives resultant amplitude $2A$, so maximum intensity scales as $(2A)^2 = 4A^2$.

   Concept and Intuition

   When two coherent waves of the same amplitude and frequency overlap, their displacements add algebraically. At points where they are exactly in phase, the amplitudes add directly, giving the largest possible resultant amplitude. Since intensity is proportional to the square of amplitude (not the amplitude itself), doubling the amplitude quadruples the intensity — interference is a much bigger multiplier on intensity than it is on amplitude.

   Step-by-Step Solution

   1. Each wave has amplitude $A$, so its own intensity is $I \propto A^2$.
   2. At a point of constructive interference (path difference = integer multiple of $\lambda$), the resultant amplitude is $A_{max} = A + A = 2A$. …

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2. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.Young's double slit experiment is conducted in water. The refractive index of water is 1.33. Wavelength of incident light is 6000 Å and the separation between the slits is 2 mm. Fringe width is given by 0.63 mm. The distance between screen and slits is (A) 1.4 m (B) 1.04 m (C) 2.79 m (D) 1.72 m

   ›Reveal solutionSolution

   This tests YDSE fringe width when the experiment is performed inside a refractive medium; the wavelength inside the medium must be used, giving $D \approx 2.79\ m$.

   Concept and Intuition

   Fringe width in Young's double-slit experiment depends on the wavelength inside the medium where the interference actually happens, not the vacuum/air wavelength. When the whole setup is submerged in water, the effective wavelength shrinks by a factor of $\mu$ (the medium's refractive index), which in turn changes the fringe width for the same slit separation and screen distance — so we must first find the water wavelength before applying the standard fringe-width relation.

   Step-by-Step Solution

   1. Wavelength in water: $\lambda' = \dfrac{\lambda}{\mu} = \dfrac{6000\ \text{Å}}{1.33} \approx 4511.3\ \text{Å} = 4.5113\times10^{-7}\ m$.
   2. Fringe width formula: $\beta = \dfrac{\lambda' D}{d}$, so $D = \dfrac{\beta d}{\lambda'}$. …

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3. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.In a Young's double slit experiment, the intensity at a point where the path difference is $\dfrac{\lambda}{6}$ ($\lambda$ being the wavelength of the light used) is I. If $I_0$ denotes the maximum intensity, $\dfrac{I}{I_0}$ is (A) $\dfrac{1}{\sqrt{2}}$ (B) $\dfrac{\sqrt{3}}{2}$ (C) $\dfrac{1}{2}$ (D) $\dfrac{3}{4}$

   ›Reveal solutionSolution

   Converts a path difference into a phase difference and applies the two-slit interference intensity law $I = I_0\cos^2(\delta/2)$; the answer is $I/I_0 = 3/4$.

   Concept and Intuition

   In YDSE, the resultant intensity at any point depends on the phase difference $\delta$ between the two coherent waves arriving there, which comes directly from the path difference $\Delta x$: $\delta = \dfrac{2\pi}{\lambda}\Delta x$. The standard two-source interference result is

   $I = 4a^2\cos^2\left(\frac{\delta}{2}\right) = I_{max}\cos^2\left(\frac{\delta}{2}\right)$

   where $I_{max}=I_0$ is the intensity at a bright fringe ($\delta=0$). So knowing the path difference in terms of $\lambda$ immediately gives the fractional intensity, without needing the individual slit intensities.

   Step-by-Step Solution

   1. Path difference given: $\Delta x = \lambda/6$.
   2. Phase difference: $\delta = \dfrac{2\pi}{\lambda}\cdot\dfrac{\lambda}{6} = \dfrac{\pi}{3}$. …

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4. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.In Young's double slit experiment, the ratio of intensities of maxima and minima in the interference experiment is 25:9. The ratio of intensities of two slits is (A) 18:3 (B) 4:1 (C) 8:1 (D) 16:1

   ›Reveal solutionSolution

   This tests the standard fringe-visibility relation between $I_{max}/I_{min}$ and the ratio of the two slit intensities. Answer: 16:1.

   Concept and Intuition

   In Young's double-slit experiment, the resultant intensity varies between $I_{max}=(\sqrt{I_1}+\sqrt{I_2})^2$ (constructive) and $I_{min}=(\sqrt{I_1}-\sqrt{I_2})^2$ (destructive). The ratio $I_{max}/I_{min}$ depends only on the ratio of the individual slit intensities, so given the fringe contrast we can back out how unequal the two slits are.

   Step-by-Step Solution

   1. $\dfrac{I_{max}}{I_{min}} = \left(\dfrac{\sqrt{I_1}+\sqrt{I_2}}{\sqrt{I_1}-\sqrt{I_2}}\right)^2 = \dfrac{25}{9}$.
   2. Taking square roots: $\dfrac{\sqrt{I_1}+\sqrt{I_2}}{\sqrt{I_1}-\sqrt{I_2}} = \dfrac{5}{3}$.
   3. Let $r = \sqrt{I_1/I_2}$. Dividing numerator and denominator by $\sqrt{I_2}$: $\dfrac{r+1}{r-1} = \dfrac{5}{3}$.
   4. Cross-multiplying: $3(r+1) = 5(r-1) \Rightarrow 3r+3 = 5r-5 \Rightarrow 8 = 2r \Rightarrow r = 4$. …

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5. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.The correct relation between fringe width ($\beta$) and distance between the slits

   (d) is (A) [FIGURE: graph of $\beta$ vs $d$ — a straight line with negative slope, decreasing] (B) [FIGURE: graph of $\beta$ vs $d$ — a curve decreasing steeply and flattening (inverse-like), concave up] (C) [FIGURE: graph of $\beta$ vs $d$ — a straight line through the origin with positive slope, increasing] (D) [FIGURE: graph of $\beta$ vs $d$ — a curve rising from the origin and flattening to a constant value]

   ›Reveal solutionSolution

   This tests the exact functional form of fringe width versus slit separation in Young's double slit experiment: $\beta = \lambda D/d$ is an inverse relationship, so its graph is a decreasing hyperbola, not a straight line.

   Concept and Intuition

   In Young's double-slit interference, the fringe width — the spacing between consecutive bright (or dark) fringes on the screen — is given by

   $\beta = \frac{\lambda D}{d},$

   where $\lambda$ is the wavelength of light, $D$ is the distance from the slits to the screen, and $d$ is the separation between the two slits. For fixed $\lambda$ and $D$, $\beta$ is inversely proportional to $d$: $\beta \propto 1/d$. This is a hyperbola-type relationship: as $d\to 0$, $\beta\to\infty$ (very widely spaced fringes), and as $d$ increases, $\beta$ falls off rapidly at first and then more gradually, asymptotically approaching (but never touching) the horizontal axis. It is emphatically not a straight line, since a straight line would mean $\beta$ is directly proportional (or linearly related) to $d$, which contradicts the $1/d$ form.

   Step-by-Step Solution

   1. Start from the standard fringe-width formula $\beta = \lambda D/d$.
   2. Treat $\lambda D$ as a constant $k$ (for fixed source wavelength and fixed screen distance), so $\beta = k/d$.
   3. This is the equation of a rectangular hyperbola in the $\beta$–$d$ plane: $\beta \cdot d = k$ (constant). …

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6. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.In Young's double slit experiment, the fringe width is $\beta$. If the distance between slit and screen is doubled and separation between slits is made ¼ of initial value, then new fringe width is (A) $4\beta$ (B) $\dfrac{\beta}{4}$ (C) $\dfrac{\beta}{8}$ (D) $8\beta$

   ›Reveal solutionSolution

   Direct application of the YDSE fringe-width formula's dependence on screen distance and slit separation. Doubling $D$ and quartering $d$ multiplies the fringe width by 8. Answer: $8\beta$.

   Concept and Intuition

   In Young's double slit experiment, the fringe width (spacing between consecutive bright or dark fringes) is

   $\beta = \frac{\lambda D}{d}$

   where $D$ is the slit-to-screen distance and $d$ is the slit separation. Physically, moving the screen farther away ($D$ increases) spreads the interference pattern out (fringes get wider), while moving the slits closer together ($d$ decreases) also spreads the pattern out (the diffracted/interfering beams overlap over a wider angular range for the same fringe order). Both changes here act in the same direction — both increase $\beta$ — so their effects multiply together rather than partially cancel.

   Step-by-Step Solution

   1. Original fringe width: $\beta = \dfrac{\lambda D}{d}$.
   2. New conditions: $D' = 2D$ (doubled), $d' = \dfrac{d}{4}$ (quartered).
   3. New fringe width: …

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7. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.Interference fringes produced by a double slit arrangement using a monochromatic light of wave length 5,890Å, have an angular fringe width $0.28^{\circ}$. If the entire arrangement is immersed in water, the new angular fringe width will be ($^{a}\mu_{\omega}=4/3$) (A) $0.24^{\circ}$ (B) $0.21^{\circ}$ (C) $0.18^{\circ}$ (D) $0.36^{\circ}$

   ›Reveal solutionSolution

   Immersing the double-slit setup in water shortens the wavelength of light by a factor $1/\mu_w$, which shrinks the angular fringe width by the same factor, from $0.28^{\circ}$ to $0.21^{\circ}$.

   Concept and Intuition

   Angular fringe width in a Young's double slit experiment is $\theta=\lambda/d$, depending only on the wavelength (in the medium the light is actually traveling through) and the slit separation $d$ (a fixed mechanical quantity, unaffected by immersion). When the whole setup (slits, screen, and the medium between them) is immersed in water, the wavelength of light in that medium shortens to $\lambda_{water}=\lambda_{air}/\mu_w$, so the fringe pattern compresses proportionally.

   Step-by-Step Solution

   1. In air: $\theta_{air}=\lambda_{air}/d = 0.28^{\circ}$.
   2. In water: $\lambda_{water}=\lambda_{air}/\mu_w$, and $d$ is unchanged, so $\theta_{water}=\lambda_{water}/d=\theta_{air}/\mu_w$. …

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8. AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
   Q.If Young's double slit experiment is done in air first and then if the experiment is conducted by immersing the apparatus in water, the fringe width (A) remains same (B) decreases (C) increases (D) Becomes zero

   ›Reveal solutionSolution

   Tests how immersing Young's double-slit setup in a medium changes fringe width — the wavelength shrinks in a denser medium, so the fringes get closer together.

   Concept and Intuition

   Fringe width in YDSE, $\beta=\lambda D/d$, is set by the wavelength of light inside the medium the light is travelling through, not its vacuum/air wavelength. Water has a refractive index greater than air, so light slows down and its wavelength shrinks proportionally ($\lambda_{medium}=\lambda_{air}/n$), which directly shrinks the fringe spacing since everything else ($D$, $d$) is unchanged by the medium.

   Step-by-Step Solution

   1. Fringe width formula: $\beta = \dfrac{\lambda D}{d}$, where $\lambda$ is the wavelength in the medium the interference is occurring in, $D$ is slit-to-screen distance, $d$ is slit separation.
   2. In air, $\lambda = \lambda_{air}$ (essentially, since $n_{air}\approx1$).
   3. When immersed in water (refractive index $n_w\approx1.33$), the wavelength becomes $\lambda_{water} = \dfrac{\lambda_{air}}{n_w}$, which is smaller than $\lambda_{air}$.
   4. $D$ and $d$ are geometric quantities of the apparatus and don't change on immersion. …

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9. AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
   Q.Young's double slit experiment setup is in such a way that, when path difference is λ, then intensity at a point is $I$. If the path difference is $\dfrac{\lambda}{4}$, then intensity at the same point is (A) $\dfrac{I}{\sqrt{2}}$ (B) $\dfrac{I}{2}$ (C) $2I$ (D) $\sqrt{2}\,I$

   ›Reveal solutionSolution

   Using the two-slit intensity formula $I(\delta)=4I_0\cos^2(\delta/2)$, calibrating with the given path-difference-$\lambda$ case shows the intensity at path difference $\lambda/4$ is $I/2$.

   Concept and Intuition

   In Young's double slit experiment, the resultant intensity at a point depends on the phase difference $\delta$ between the two interfering waves, via $I=4I_0\cos^2(\delta/2)$, where $I_0$ is the intensity due to a single slit alone and $\delta=\dfrac{2\pi}{\lambda}\times\Delta$ for a path difference $\Delta$. A path difference of a full wavelength ($\Delta=\lambda$) gives $\delta=2\pi$, which is constructive interference (maximum intensity), letting us calibrate $I_0$ from the given information.

   Step-by-Step Solution

   1. At path difference $\Delta=\lambda$: phase $\delta=\dfrac{2\pi}{\lambda}\times\lambda=2\pi$.
   2. $I = 4I_0\cos^2(2\pi/2)=4I_0\cos^2(\pi)=4I_0(1)^2=4I_0$. Given this equals $I$: $I_0=I/4$.
   3. At path difference $\Delta=\lambda/4$: phase $\delta=\dfrac{2\pi}{\lambda}\times\dfrac{\lambda}{4}=\dfrac{\pi}{2}$. …

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10. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
    Q.In Young's double slit experiment, the wavelengths of red and blue lights used are $7.5\times10^{-5}$ cm and $5\times10^{-5}$ cm respectively. If $n^{th}$ bright fringe of red color coincides with $(n+1)^{th}$ bright fringe of blue colour, then the value of 'n' is (A) 1 (B) 2 (C) 4 (D) 8

    ›Reveal solutionSolution

    This tests fringe coincidence in Young's double-slit experiment for two wavelengths. Answer: $n=2$.

    Concept and Intuition

    In YDSE the bright fringe positions for a given wavelength form an evenly-spaced set, $y_n=\dfrac{n\lambda D}{d}$. Two different colours have different fringe spacings, so their bright fringes drift apart and occasionally land at the same point on the screen — this is fringe coincidence. At the coincidence point, the physical position $y$ must be identical for both colours, even though the fringe order $n$ differs.

    Step-by-Step Solution

    1. Position of $n^{th}$ red bright fringe: $y=\dfrac{n\lambda_{red}D}{d}$.
    2. Position of $(n+1)^{th}$ blue bright fringe: $y=\dfrac{(n+1)\lambda_{blue}D}{d}$.
    3. Equate (same $D,d$, so they cancel): $n\lambda_{red}=(n+1)\lambda_{blue}$. …

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11. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
    Q.In Young's double slit experiment, the wavelength of monochromatic light is increased by 20% and the distance between the two slits is decreased by 25%. If the initial fringe width is 0.3 mm, then the final fringe width is (A) 0.72 mm (B) 0.60 mm (C) 0.16 mm (D) 0.48 mm

    ›Reveal solutionSolution

    Fringe width in YDSE scales directly with wavelength and inversely with slit separation. Increasing $\lambda$ by 20% and decreasing $d$ by 25% together multiply the fringe width by 1.6, giving 0.48 mm.

    Concept and Intuition

    In Young's double slit experiment, bright fringes form where path difference is an integer multiple of $\lambda$; the spacing between adjacent fringes on the screen is $\beta = \lambda D/d$. Increasing the wavelength spreads the fringes further apart (a 'wider' interference pattern), while increasing slit separation squeezes them closer together — hence $\beta$ scales directly with $\lambda$ and inversely with $d$. Both changes here work in the SAME direction (increasing $\beta$): a longer wavelength directly increases it, and a smaller slit separation ALSO increases it.

    Step-by-Step Solution

    1. Fringe width formula: $\beta = \dfrac{\lambda D}{d}$.
    2. New wavelength: $\lambda' = \lambda + 20\%\lambda = 1.2\lambda$.
    3. New slit separation: $d' = d - 25\%d = 0.75d$. …

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12. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
    Q.In Young's double slit experiment, the distance between the slits is 0.2 cm, the distance between the screen and the slits is 1 m. If the wavelength of light used in the experiment is 5000 Å, then the distance between two consecutive dark fringes on the screen is (A) 0.25 mm (B) 0.26 mm (C) 0.27 mm (D) 0.28 mm

    ›Reveal solutionSolution

    The spacing between consecutive dark (or bright) fringes in YDSE equals the fringe width $\beta=\lambda D/d$. Answer: 0.25 mm.

    Concept and Intuition

    In Young's double slit experiment, both bright and dark fringes are equally spaced (they alternate), so the distance between any two consecutive dark fringes is the same as the fringe width $\beta$, which depends on the wavelength of light, the slit separation, and the screen distance.

    Step-by-Step Solution

    1. Convert values to SI: $d = 0.2\ \text{cm} = 2\times10^{-3}\ \text{m}$; $D = 1\ \text{m}$; $\lambda = 5000\ \text{Å} = 5000\times10^{-10}\ \text{m} = 5\times10^{-7}\ \text{m}$.
    2. Fringe width: $\beta = \dfrac{\lambda D}{d} = \dfrac{5\times10^{-7}\times1}{2\times10^{-3}}$.
    3. Compute: $\beta = 2.5\times10^{-4}\ \text{m} = 0.25\ \text{mm}$. …

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