Question

In the Young's double slit experiment the fringe width of the interference pattern is found to be \(3.2 \times 10^{-4}\) m, when the light of wavelength 6400 \(\text{Å}\) is use(D) What will be the change in fringe width if the light is replaced with a light of wavelength 4800 \(\text{Å}\)?

• A. \( 2.4 \times 10^{-4} \, \text{m} \)
• B. \( 1.6 \times 10^{-4} \, \text{m} \)
• C. \( 0.8 \times 10^{-4} \, \text{m} \)
• D. \( 5.6 \times 10^{-4} \, \text{m} \)

Hint:

In Young's double-slit experiment, the fringe width is directly proportional to the wavelength of the light use(D) If the wavelength decreases, the fringe width also decreases.

Answer 1

The Correct Option is C

In Young's double-slit experiment, the fringe width \( \beta \) is given by the formula: \[ \beta = \frac{\lambda D}{d} \] where: - \( \lambda \) is the wavelength of the light used, - \( D \) is the distance between the screen and the slits, - \( d \) is the distance between the slits. Now, when the wavelength changes, the fringe width changes accordingly. Let the initial wavelength \( \lambda_1 = 6400 \, \text{Å} = 6400 \times 10^{-10} \, \text{m} \) and the initial fringe width \( \beta_1 = 3.2 \times 10^{-4} \, \text{m} \). The new wavelength is \( \lambda_2 = 4800 \, \text{Å} = 4800 \times 10^{-10} \, \text{m} \). We can use the ratio of the fringe widths to find the new fringe width \( \beta_2 \): \[ \frac{\beta_2}{\beta_1} = \frac{\lambda_2}{\lambda_1} \] Substitute the values: \[ \frac{\beta_2}{3.2 \times 10^{-4}} = \frac{4800 \times 10^{-10}}{6400 \times 10^{-10}} \] Simplifying the right-hand side: \[ \frac{\beta_2}{3.2 \times 10^{-4}} = \frac{4800}{6400} = 0.75 \] Now, solve for \( \beta_2 \): \[ \beta_2 = 3.2 \times 10^{-4} \times 0.75 = 0.8 \times 10^{-4} \, \text{m} \] Thus, the new fringe width is \( 0.8 \times 10^{-4} \, \text{m} \).