JEE Challenger
More from Wave Optics
PhyMedAdv 2014 • Paper 1

Fringe Width and Interference Patterns in Bichromatic Young Double Slit

A light source emitting two wavelengths λ1=400 nm\lambda_1 = 400\text{ nm} and λ2=600 nm\lambda_2 = 600\text{ nm} is employed in a Young's double-slit experiment. Let the recorded fringe widths corresponding to λ1\lambda_1 and λ2\lambda_2 be β1\beta_1 and β2\beta_2, and the number of fringes observed within a given distance yy on one side of the central bright fringe be m1m_1 and m2m_2, respectively. Which of the following statements is/are correct?

Options

A

β2>β1\beta_2 > \beta_1

Correct
B

m1>m2m_1 > m_2

Correct
C

Measured from the central maximum, the 3rd3^{\text{rd}} maximum of λ2\lambda_2 coincides with the 5th5^{\text{th}} minimum of λ1\lambda_1

Correct
D

The angular separation of fringes for λ1\lambda_1 is greater than that for λ2\lambda_2

Topics & Concepts

Wave OpticsYoung's Double SlitInterference

Step-by-Step Solution

To determine the correct statements, let us analyze each option step-by-step:

  1. Fringe Width (β\beta): The fringe width in Young's double-slit experiment is given by the formula: β=λDd\beta = \frac{\lambda D}{d} where:

    • λ\lambda is the wavelength of light,
    • DD is the distance between the slits and the screen,
    • dd is the distance between the two slits.

    Given λ1=400 nm\lambda_1 = 400\text{ nm} and λ2=600 nm\lambda_2 = 600\text{ nm}: λ2>λ1  ⟹  β2>β1\lambda_2 > \lambda_1 \implies \beta_2 > \beta_1 Thus, Option (A) is correct.

  2. Number of Fringes within a Given Distance yy (mm): The number of fringes mm observed within a distance yy on the screen is given by: m=yβ=y⋅dλDm = \frac{y}{\beta} = \frac{y \cdot d}{\lambda D} Since m∝1λm \propto \frac{1}{\lambda}: λ1<λ2  ⟹  m1>m2\lambda_1 < \lambda_2 \implies m_1 > m_2 Thus, Option (B) is correct.

  3. Coincidence of Maxima and Minima:

    • The position of the nthn^{\text{th}} maximum on the screen is: yn,max=nλDdy_{n, \text{max}} = n \frac{\lambda D}{d} For the 3rd3^{\text{rd}} maximum of λ2\lambda_2 (n=3n = 3): y3,max(λ2)=3⋅600×10−9Dd=1800×10−9Ddy_{3, \text{max}}(\lambda_2) = 3 \cdot \frac{600 \times 10^{-9} D}{d} = 1800 \times 10^{-9} \frac{D}{d}

    • The position of the nthn^{\text{th}} minimum on the screen is: yn,min=(n−12)λDdy_{n, \text{min}} = \left(n - \frac{1}{2}\right) \frac{\lambda D}{d} For the 5th5^{\text{th}} minimum of λ1\lambda_1 (n=5n = 5): y5,min(λ1)=(5−12)400×10−9Dd=4.5×400×10−9Dd=1800×10−9Ddy_{5, \text{min}}(\lambda_1) = \left(5 - \frac{1}{2}\right) \frac{400 \times 10^{-9} D}{d} = 4.5 \times 400 \times 10^{-9} \frac{D}{d} = 1800 \times 10^{-9} \frac{D}{d}

    Since y3,max(λ2)=y5,min(λ1)y_{3, \text{max}}(\lambda_2) = y_{5, \text{min}}(\lambda_1), the 3rd3^{\text{rd}} maximum of λ2\lambda_2 overlaps with the 5th5^{\text{th}} minimum of λ1\lambda_1. Thus, Option (C) is correct.

  4. Angular Separation (θ\theta): The angular separation between adjacent fringes is given by: θ=βD=λd\theta = \frac{\beta}{D} = \frac{\lambda}{d} Since λ1<λ2\lambda_1 < \lambda_2: θ1<θ2\theta_1 < \theta_2 Therefore, the angular separation of fringes for λ1\lambda_1 is less than that for λ2\lambda_2. Thus, Option (D) is incorrect.

Conclusion: The correct statements are (A), (B), and (C).

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