In Young's double slit experiment, slits $S_1$ and $S_2$ are separated by a distance $d$, and a screen is placed at a distance $D\gg d$ from the slits. Derive an expression for the path difference at a point $P$ on the screen at a distance $y$ from the centre, and use it to obtain the fringe width $\beta$.

In Young's double slit experiment, slits S₁ and S₂ are separated by a distance d, and a screen is placed at a

Example · Example 6

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Let $P$ be a point on the screen at height $y$ above the central point $O$ . Since $D\gg d$ , the lines $S_1P$ and $S_2P$ are very nearly parallel, both making nearly the same small angle with the central axis. Dropping a perpendicular from $S_1$ onto the line $S_2P$ marks off the extra path $S_2P-S_1P$ as one side of a small right triangle whose hypotenuse is $S_1S_2=d$ and whose angle at $S_1$ (or $S_2$ ) equals the angle $\theta$ that $OP$ makes with the central axis at the slits, where $\tan\theta=y/D\approx\theta$ for small angles. This gives

\Delta=S_2P-S_1P\approx d\sin\theta\approx d\theta=\frac{yd}{D}

For the $n$ th bright fringe, the constructive-interference condition $\Delta=n\lambda$ then gives its position:

$\frac{y_nd}{D}=n\lambda\ \Rightarrow\ y_n=\frac{n\lambda D}{d}$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

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CBSE 2026Set ANNUAL1 mark
$Q. Answer in one word/sentence: If red light source is used instead of blue light in Young's double slit experiment, what happens with Fringe width?$
›Reveal solutionSolution

Fringe width is directly proportional to wavelength, and red light has a longer wavelength than blue, so the fringe width increases.

…

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CBSE 2026Set ANNUAL1 mark
$5\times10^{-5}$
›Reveal solutionSolution

Using the standard Young's double-slit fringe-width formula $\beta=\lambda D/d$ with the given wavelength, slit separation, and screen distance gives a fringe width of exactly $1$ mm.

Formula

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

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

where $\lambda$ = wavelength of light, $D$ = distance from the slits to the screen, and $d$ = separation between the two slits.

Converting the given quantities to SI units

$\lambda = 5\times10^{-5}\text{ cm} = 5\times10^{-7}\text{ m}$

$d = 0.025\text{ mm} = 2.5\times10^{-5}\text{ m}$ …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. In a Young's double-slit experiment, the slit separation is doubled. To maintain the same fringe spacing on the screen, the screen-to-slit distance D must be changed to :$
(a) $\sqrt{2}D$
(b) $\dfrac{D}{2}$
(c) $\dfrac{D}{\sqrt{2}}$
(d) $2D$
›Reveal solutionSolution

Since fringe width $\beta\propto D/d$ , doubling the slit separation $d$ requires doubling the screen distance $D$ to keep the fringe spacing the same.

Working

Fringe width in Young's double-slit experiment: $\beta = \dfrac{\lambda D}{d}$ .

Let $d'=2d$ . For $\beta'=\beta$ : …

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CBSE 2025Set ANNUAL1 markMCQ
$Q.In Young's double slit experiment the fringe width is found to be β. If the entire apparatus is immersed in liquid of refractive index n, the new fringe width will be :$
(a) β
(b) nβ
(c) β/n
(d) n²β
›Reveal solutionSolution

Immersing the YDSE setup in a liquid of refractive index n shrinks the wavelength of light inside the liquid, so the fringe width shrinks by the same factor n.

Fringe width in air is

$\beta = \frac{\lambda D}{d}$ …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. How is interference pattern in double slit experiment affected, if a source of blue light is used in place of yellow light producing the same intensity ?$
(a) The fringe width will decrease
(b) The fringe width will increase
(c) The fringe width will become brighter
(d) The fringe width will become fainter
›Reveal solutionSolution

Blue light has a shorter wavelength than yellow light, and fringe width is directly proportional to wavelength.

In Young's double slit experiment, fringe width is given by

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

…

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CBSE 2024Set 55/5/11 markMCQ
$Q. Assertion (A): In Young's double slit experiment, when the two coherent sources are infinitely close to each other, the interference pattern cannot be observed. Reason (R): The fringe width is proportional to the separation between the two sources. (A) Both A and R are true and R is the correct explanation of A. (B) Both A and R are true but R is NOT the correct explanation of A. (C) A is true but R is false. (D) Both A and R are false.$
›Reveal solutionSolution

As the two coherent sources are brought infinitely close ( $d\to0$ ), the fringe width $\beta=\lambda D/d\to\infty$ -- the entire screen falls inside a single fringe, so no observable interference pattern forms. Assertion (A) is true. But Reason (R), as stated, claims fringe width is proportional to separation $d$ -- the real relation $\beta=\lambda D/d$ makes $\beta$ inversely proportional to $d$ , so R (as worded) is false. Option (C).

Fringe width and source separation

In Young's double-slit experiment, the fringe width (spacing between consecutive bright or dark fringes) is
$\beta=\frac{\lambda D}{d},$
where $\lambda$ is the wavelength, $D$ the screen distance, and $d$ the separation between the two coherent sources.
$\beta=\frac{\lambda D}{d}\quad\Rightarrow\quad \beta\ \text{is INVERSELY proportional to }d,\ \text{not directly proportional.}$
Checking Assertion (A). As $d\to0$ , $\beta=\lambda D/d\to\infty$ : the fringes become infinitely wide, so within any real, finite screen only a single bright region is seen -- no distinguishable interference pattern (alternating bright/dark fringes) can be observed. So A is physically correct. …
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CBSE 2024Set A1 markMCQ
$Q. The fringe width in interference of light due to two coherent sources is (A) proportional to wavelength (B) inversely proportional to wavelength (C) proportional to square of wavelength (D) inversely proportional to square of wavelength$
›Reveal solutionSolution

In Young's double-slit experiment fringe width β = λD/d, hence β is directly proportional to wavelength → option (A).

The fringe width in a two-source interference pattern is given by

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

…

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CBSE 2024Set ANNUAL1 markMCQ
$5\times10^{-5}$
(a) 0.1 mm
(b) 1 mm
(c) 0.01 mm
(d) None of the above
›Reveal solutionSolution

In Young's double-slit experiment the fringe width is $\beta = \lambda D/d$ , where $\lambda$ is the wavelength, $D$ the slit-to-screen distance and $d$ the slit separation. Substituting the given values directly gives the fringe width.

Convert every quantity to SI units

$\lambda = 5\times10^{-5}\,\text{cm} = 5\times10^{-5}\times10^{-2}\,\text{m} = 5\times10^{-7}\,\text{m}$

$d = 0.025\,\text{mm} = 0.025\times10^{-3}\,\text{m} = 2.5\times10^{-5}\,\text{m}$

$D = 5\,\text{cm} = 5\times10^{-2}\,\text{m}$

Apply the fringe-width formula

…

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CBSE 2023Set F1 markMCQ
$Q. When the distance between source of light and screen is increased, then fringe width (A) increases (B) decreases (C) remains same (D) none of these$
›Reveal solutionSolution

$\beta = \dfrac{\lambda D}{d}$ ; increasing D (screen distance) increases the fringe width.

In Young's double-slit experiment the fringe width is

$\beta = \frac{\lambda D}{d},$ …

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CBSE 2023Set ANNUAL1 mark
$Q. In Young's double-slit experiment, what happens to the fringe width if the distance between the slits is increased? (Write the answer only)$
›Reveal solutionSolution

In Young's double-slit experiment, fringe width $\beta = \lambda D/d$ is inversely proportional to the slit separation d, so increasing d decreases the fringe width.

The fringe width in YDSE is given by

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

…

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CBSE 2023Set ANNUAL1 markMCQ
$Q. In Young's double-slit experiment, the slit separation is doubled. To maintain the same fringe spacing on the screen, the screen-to-slit distance D must be changed to :$
(a) $\sqrt{2}D$
(b) $2D$
(c) $\dfrac{D}{\sqrt{2}}$
(d) $\dfrac{D}{2}$
›Reveal solutionSolution

Since fringe width $\beta \propto D/d$ , doubling the slit separation $d$ requires doubling the screen distance $D$ to keep $\beta$ unchanged.

Working

In Young's double-slit experiment, the fringe width is

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

…

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CBSE 2022Set ANNUAL1 mark
$Q. What will be the effect on fringe width in Young's double-slit experiment, if the distance between slit and screen increased?$
›Reveal solutionSolution

Since β = λD/d, increasing D increases the fringe width proportionally, while λ and d are unchanged.

In Young's double-slit experiment, the fringe width (distance between two consecutive bright or dark fringes) is given by

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

where λ is the wavelength of light used, D is the distance between the plane of the slits and the screen, and d is the separation between the two slits.

…

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