Question

The fringe width for red colour as compare to that for violet colour is approximately

• A. 4 times
• B. 3 times
• C. 8 times
• D. double

Answer 1

The Correct Option is D

Fringe Width in Double-Slit Interference

The fringe width in a double-slit interference pattern is given by the formula:

\[ w = \frac{\lambda L}{d} \]

Here, the wavelength \(\lambda\) is directly proportional to the fringe width \(w\).

- Red light has a wavelength of approximately 700 nm.
- Violet light has a wavelength of approximately 400 nm.

To compare the fringe widths:

\[ \frac{w_r}{w_v} = \frac{\lambda_r}{\lambda_v} \]

Substituting values:

\[ \frac{w_r}{w_v} \approx \frac{700}{400} \approx 1.75 \]

This means the fringe width for red light is about 1.75 times that of violet light.

Since none of the given answer choices (A, B, C, D) match this exactly, the closest option is (D) "double", suggesting that the fringe width of red light is nearly twice that of violet light.

Answer 2

The formula for fringe width (\( \beta \)) in interference experiments like Young's Double Slit Experiment is given by: \[ \beta = \frac{\lambda D}{d} \] where: 

• \( \lambda \) is the wavelength of the light used.
• \( D \) is the distance from the slits (or sources) to the screen.
• \( d \) is the separation between the slits (or sources).

 

From this formula, we can see that the fringe width is directly proportional to the wavelength of light, assuming \( D \) and \( d \) are kept constant: \[ \beta \propto \lambda \]

We need to compare the fringe width for red light (\( \beta_{\text{red}} \)) with the fringe width for violet light (\( \beta_{\text{violet}} \)). This depends on the ratio of their wavelengths (\( \lambda_{\text{red}} \) and \( \lambda_{\text{violet}} \)).

The visible light spectrum ranges from violet to red.

• The wavelength of violet light (\( \lambda_{\text{violet}} \)) is approximately 400 nm (\( 4 \times 10^{-7} \) m).
• The wavelength of red light (\( \lambda_{\text{red}} \)) is approximately 700 nm - 800 nm (\( 7 \times 10^{-7} \) m to \( 8 \times 10^{-7} \) m).

 

Let's calculate the ratio of the fringe widths: \[ \frac{\beta_{\text{red}}}{\beta_{\text{violet}}} = \frac{\lambda_{\text{red}}}{\lambda_{\text{violet}}} \] Using typical approximate values: \[ \frac{\beta_{\text{red}}}{\beta_{\text{violet}}} \approx \frac{700 \text{ nm}}{400 \text{ nm}} = \frac{7}{4} = 1.75 \] If we consider the upper range for red light: \[ \frac{\beta_{\text{red}}}{\beta_{\text{violet}}} \approx \frac{800 \text{ nm}}{400 \text{ nm}} = \frac{8}{4} = 2 \]

Therefore, the fringe width for red light is approximately 1.75 to 2 times the fringe width for violet light. Comparing this ratio to the given options:

• 4 times
• 3 times
• 8 times
• double

The closest and most appropriate description is that the fringe width for red colour is approximately double that for violet colour.