Question

In Young's double-slit experiment, two different light beams of wavelengths $λ_1$ and $λ_2$ produce interference pattern with band widths $β_1$ and $β_2$ respectively. If the ratio between $β_1$ and $β_2$ is 3:2, then the ratio between and is

• A. 3:1
• B. 1:3
• C. 2:3
• D. 4:5
• E. 3:2

Answer 1

Answer: (E)

Young's Double-Slit Experiment: Bandwidth and Wavelength Ratio

In Young's double-slit experiment, two different light beams of wavelengths \(\lambda_1\) and \(\lambda_2\) produce interference patterns with band widths \(\beta_1\) and \(\beta_2\) respectively. We are given that the ratio between \(\beta_1\) and \(\beta_2\) is 3:2, and we need to find the ratio between \(\lambda_1\) and \(\lambda_2\). 

Step 1: Bandwidth Formula

The bandwidth (fringe width) \(\beta\) in Young's double-slit experiment is given by:

\(\beta = \frac{\lambda D}{d}\)

Where:

• \(\beta\) is the bandwidth (fringe width).
• \(\lambda\) is the wavelength of the light.
• \(D\) is the distance from the slits to the screen.
• \(d\) is the separation between the slits.

Step 2: Ratio of Bandwidths

We are given that \(\frac{\beta_1}{\beta_2} = \frac{3}{2}\). We can write this as:

\(\frac{\beta_1}{\beta_2} = \frac{\frac{\lambda_1 D}{d}}{\frac{\lambda_2 D}{d}}\)

Step 3: Simplify the Ratio

The terms \(D\) and \(d\) are the same for both wavelengths (since it's the same experimental setup), so they cancel out:

\(\frac{\beta_1}{\beta_2} = \frac{\lambda_1}{\lambda_2}\)

Step 4: Find the Ratio of Wavelengths

Since \(\frac{\beta_1}{\beta_2} = \frac{3}{2}\), we have:

\(\frac{\lambda_1}{\lambda_2} = \frac{3}{2}\)

Conclusion

The ratio between \(\lambda_1\) and \(\lambda_2\) is 3:2.

Answer 2

In Young's Double-Slit Experiment, the fringe width \( \beta \) of the interference pattern 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 slits and the screen, - \( d \) is the distance between the two slits. 

Suppose two light sources with different wavelengths \( \lambda_1 \) and \( \lambda_2 \) produce fringe widths \( \beta_1 \) and \( \beta_2 \) respectively. Then, \[ \beta_1 = \frac{\lambda_1 D}{d}, \quad \beta_2 = \frac{\lambda_2 D}{d} \]

Now taking the ratio of the two fringe widths: \[ \frac{\beta_1}{\beta_2} = \frac{\lambda_1 D / d}{\lambda_2 D / d} = \frac{\lambda_1}{\lambda_2} \] This implies: \[ \frac{\beta_1}{\beta_2} = \frac{\lambda_1}{\lambda_2} \]

Given: \[ \frac{\beta_1}{\beta_2} = \frac{3}{2} \Rightarrow \frac{\lambda_1}{\lambda_2} = \frac{3}{2} \]

Final Answer: \( \boxed{3:2} \)