Q125 / 180 · COMEDK 2023 (Evening Shift) · Physics

In the young's double slit experiment the fringe width of the interference pattern is found to be $3.2 \times 10^{-4} \mathrm{~m}$, when the light of wave length $6400^{\circ} \mathrm{A}$ is used. What will be change in fringe width if the light is replaced with a light of wave length $4800^{\circ} \mathrm{A}$

  • A
    $1.6 \times 10^{-4} \mathrm{~m}$
  • B
    $2.4 \times 10^{-4} \mathrm{~m}$
  • C
    $5.6 \times 10^{-4} \mathrm{~m}$
  • Correct answer
    D
    $0.8 \times 10^{-4} \mathrm{~m}$

Solution

The fringe width $\beta$ in Young's double slit experiment is given by the formula $\beta = \dfrac{\lambda D}{d}$, where $\lambda$ is the wavelength of light, $D$ is the distance between the screen and the slits, and $d$ is the distance between the two slits.

Given the initial state:
$\beta_1 = 3.2 \times 10^{-4} \text{ m}$
$\lambda_1 = 6400 \text{ \AA}$

For the second state:
$\lambda_2 = 4800 \text{ \AA}$

Since $\beta \propto \lambda$, we have $\dfrac{\beta_2}{\beta_1} = \dfrac{\lambda_2}{\lambda_1}$.

Substituting the values:
$\beta_2 = \beta_1 \times \dfrac{\lambda_2}{\lambda_1} = 3.2 \times 10^{-4} \times \dfrac{4800}{6400}$

$\beta_2 = 3.2 \times 10^{-4} \times \dfrac{3}{4} = 2.4 \times 10^{-4} \text{ m}$

The change in fringe width is $\Delta \beta = \beta_1 - \beta_2 = 3.2 \times 10^{-4} - 2.4 \times 10^{-4} = 0.8 \times 10^{-4} \text{ m}$.

Answer: $0.8 \times 10^{-4} \mathrm{~m}$

Hence, the correct answer is option: 'D'.

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