Derive an expression for bandwidth of interference fringes in Young's double slit experiment.

Question 154 of 199

Q. Derive an expression for bandwidth of interference fringes in Young's double slit experiment.

Tamil Nadu DgeTamil Nadu HSC (DGE) Board 2017Subjective· 10mImportance★★★★★

77% · 154/199 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Computing the path difference at a general point on the screen in Young's double slit setup and applying the conditions for bright and dark fringes gives the positions of successive fringes, whose spacing (the bandwidth) works out to $\beta=\lambda D/d$ .

Setup

Two narrow, coherent slits $S_1$ and $S_2$ , separated by a small distance $d$ , are illuminated by monochromatic light of wavelength $\lambda$ from a single source, so that $S_1$ and $S_2$ act as coherent sources. A screen is placed parallel to the plane of the slits at a distance $D$ from them, with $D \gg d$ . Let $O$ be the point on the screen on the perpendicular bisector of $S_1S_2$ , and let $P$ be a point on the screen at distance $x$ from $O$ .

Path difference at P

Using the right-angled triangles formed by $S_1$ , $S_2$ , and $P$ (with $S_1S_2=d$ and the perpendicular screen distance $D$ ):

$S_2P^2 - S_1P^2 = \left[D^2+\left(x+\frac{d}{2}\right)^2\right] - \left[D^2+\left(x-\frac{d}{2}\right)^2\right] = 2xd$

so $(S_2P-S_1P)(S_2P+S_1P) = 2xd$ . Since $D\gg d$ and $D\gg x$ , we may approximate $S_2P+S_1P \approx 2D$ , giving the path difference:

$\delta = S_2P - S_1P = \frac{2xd}{2D} = \frac{xd}{D}$

Condition for bright and dark fringes

Bright fringe (constructive interference): occurs when the path difference is an integral multiple of the wavelength:

$\delta = n\lambda \quad (n=0,\pm1,\pm2,\ldots) \Rightarrow \frac{x_n d}{D} = n\lambda \Rightarrow x_n = \frac{n\lambda D}{d}$

Dark fringe (destructive interference): occurs when the path difference is an odd multiple of $\lambda/2$ :

$\delta = \left(n+\frac{1}{2}\right)\lambda \Rightarrow x_n' = \left(n+\frac{1}{2}\right)\frac{\lambda D}{d}$

Fringe width (bandwidth)

…

Start your 14-day free trial to unlock the full solution →

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

Showing the 12 most recent of 17 on this concept.

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.

…

Start your 14-day free trial to see the full solution →
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

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

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

Start your 14-day free trial to see the full solution →
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 $\beta'=\beta$ : …

Start your 14-day free trial to see the full solution →
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

$\beta = \frac{\lambda D}{d}$ …

Start your 14-day free trial to see the full solution →
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

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

…

Start your 14-day free trial to see the full solution →
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. …
Start your 14-day free trial to see the full solution →
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

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

…

Start your 14-day free trial to see the full solution →
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

…

Start your 14-day free trial to see the full solution →
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

$\beta = \frac{\lambda D}{d},$ …

Start your 14-day free trial to see the full solution →
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

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

…

Start your 14-day free trial to see the full solution →
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

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

…

Start your 14-day free trial to see the full solution →
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.

…

Start your 14-day free trial to see the full solution →

←Previous questionQ5 — State and explain Brewster's law.→Next questionQ7 — A diffraction pattern is obtained using a beam of red light. What happ…

Q.Derive an expression for bandwidth of interference fringes in Young's double slit experiment.

Concept understanding — Fringe Width in Young's Double Slit Experiment

Setup

Path difference at P

Condition for bright and dark fringes

Fringe width (bandwidth)

Fringe width and source separation

Q.Derive an expression for bandwidth of interference fringes in Young's double slit experiment.

Setup

Path difference at P

Condition for bright and dark fringes

Fringe width (bandwidth)

🎓Unlock everything free for 14 days

Unlock everything free for 14 days