Q.In a Young's double slit arrangement with slit separation $d=0.5\ \text{mm}$ and screen distance $D=1.2\ \text{m}$, the 4th bright fringe from the centre is observed at a distance of $4.8\ \text{mm}$. Calculate the wavelength of the light used.

Showing the 12 most recent of 17 on this concept.

1. 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.

   Fringe width in Young's double slit experiment: $\beta = \dfrac{\lambda D}{d}$, where $\lambda$ is the wavelength, $D$ is the screen distance, and $d$ is the slit separation. Since $\lambda_{red} > \lambda_{blue}$, replacing blue light with red light increases $\beta$.

   ✓Final answer

   Fringe width increases when red light is used instead of blue.

2. CBSE 2026Set ANNUAL1 mark
   Q.What is the fringe width in Young's double-slit experiment if a monochromatic ray of wavelength $5\times10^{-5}$ cm falls on a double-slit separated by a distance of 0.025 mm and the screen is held at a distance of 5 cm?

   ›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

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

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

   $D = 5\text{ cm} = 5\times10^{-2}\text{ m} = 0.05\text{ m}$

   Substituting

   $\beta=\frac{(5\times10^{-7})(0.05)}{2.5\times10^{-5}}=\frac{2.5\times10^{-8}}{2.5\times10^{-5}}=1\times10^{-3}\text{ m}$

   $\beta = 1\times10^{-3}\text{ m} = 1\text{ mm}$

   ✓Final answer

   The fringe width is $\beta = 1\text{ mm}\ (=0.1\text{ cm})$.

3. 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 $\beta'=\beta$:

   $\dfrac{\lambda D'}{2d} = \dfrac{\lambda D}{d}$

   $D' = 2D$

   ✓Final answer

   The correct option is (d): the screen distance must become $2D$

4. 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

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

   When immersed in a medium of refractive index n, the wavelength inside the medium becomes $\lambda' = \lambda/n$ (frequency stays the same, speed decreases). D and d (geometry of the setup) are unaffected. So the new fringe width is

   $\beta' = \frac{\lambda' D}{d} = \frac{(\lambda/n)D}{d} = \frac{\beta}{n}$

   ✓Final answer

   (c) β/n.

5. 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

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

   where $D$ is the slit-to-screen distance and $d$ is the slit separation (both unchanged here). Blue light has a shorter wavelength ($\lambda_{blue} \approx 450$–$495$ nm) than yellow light ($\lambda_{yellow} \approx 570$–$590$ nm). Since $\beta \propto \lambda$, using blue light in place of yellow light (same source intensity) makes the fringe width smaller — the fringes get closer together.

   ✓Final answer

   The fringe width will decrease (option a).

6. 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.

   Checking Reason (R). R states the fringe width is proportional to the separation $d$. From $\beta=\lambda D/d$, $\beta$ actually varies as $1/d$ -- it is inversely proportional to $d$, the opposite of what R claims. So R, exactly as worded, is false.

   Watch out

   A common mistake is to accept R just because it "sounds related" to A. The physics of A is genuinely explained by the inverse relationship $\beta\propto 1/d$ -- R states the relationship backwards, so it is false even though A itself is true.

   ✓Final answer

   (C) A is true but R is false.

7. 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

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

   where λ is the wavelength, D the slit-to-screen distance and d the slit separation. For a fixed D and d, the fringe width is directly proportional to the wavelength of the light used. That is why red fringes are wider than blue fringes.

   ✓Final answer

   (A) proportional to wavelength.

8. CBSE 2024Set ANNUAL1 markMCQ
   Q.In Young's double-slit experiment, a monochromatic ray of light of wavelength $5\times10^{-5}$ cm falls on a double-slit of slit width $0.025$ mm. If the phenomenon of interference is observed on the screen at a distance of 5 cm, the fringe width becomes

   (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

   $\beta = \frac{\lambda D}{d} = \frac{(5\times10^{-7}\,\text{m})(5\times10^{-2}\,\text{m})}{2.5\times10^{-5}\,\text{m}} = \frac{25\times10^{-9}}{2.5\times10^{-5}}\,\text{m} = 1\times10^{-3}\,\text{m}$

   $\beta = 1\,\text{mm}$

   ✓Final answer

   (b) 1 mm

9. 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

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

   where $D$ is the distance from the slits to the screen and $d$ the slit separation. As the source/screen distance $D$ increases, $\beta$ increases in direct proportion — the fringes spread out.

   ✓Final answer

   (A) increases.

10. 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

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

    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. Since $\beta \propto 1/d$, increasing the slit separation $d$ (keeping $\lambda$ and $D$ fixed) makes the fringes narrower, i.e. the fringe width decreases and the fringes crowd closer together.

    ✓Final answer

    The fringe width decreases as the slit separation is increased.

11. 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

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

    Let the new slit separation be $d' = 2d$. For the fringe spacing to remain the same ($\beta' = \beta$):

    $\dfrac{\lambda D'}{2d} = \dfrac{\lambda D}{d}$

    $D' = 2D$

    ✓Final answer

    The correct option is (b): the screen distance must be changed to $2D$

12. 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.

    From this relation, β ∝ D (for fixed λ and d). So if the distance between the slits and the screen is increased, the fringe width increases proportionally — the fringes become wider and more widely spaced.

    ✓Final answer

    Fringe width β increases proportionally with D, since β = λD/d.