Question

A double slit interference experiment performed with a light of wavelength 600 nm forms an interference fringe pattern on a screen with 10th bright fringe having its centre at a distance of 10 mm from the central maximum. Distance of the centre of the same 10th bright fringe from the central maximum when the source of light is replaced by another source of wavelength 660 nm would be:

Hint:

In interference patterns, the fringe separation depends on the wavelength of light used. A longer wavelength results in a greater fringe separation.

Answer 1

Correct Answer: 11

Step 1: Formula for Position of Bright Fringe

In a double-slit interference pattern, the position of the \( n \)-th bright fringe is given by: \[ y_n = \frac{n \lambda D}{d} \] Where:

• \( y_n \) = Position of the \( n \)-th bright fringe
• \( \lambda \) = Wavelength of light
• \( D \) = Distance between slits and screen
• \( d \) = Distance between the slits

Step 2: Given Data

For wavelength \( \lambda_1 = 600 \) nm and the 10th bright fringe: \[ y_{10} = 10 \text{ mm} \]

For wavelength \( \lambda_2 = 660 \) nm, we need to find the new position of the 10th bright fringe.

Step 3: Finding the New Position

From the formula, the position of the fringe is directly proportional to the wavelength of light. \[ \frac{y_{10}'}{y_{10}} = \frac{\lambda_2}{\lambda_1} \] \[ y_{10}' = y_{10} \times \frac{\lambda_2}{\lambda_1} \] \[ y_{10}' = 10 \times \frac{660}{600} \] \[ y_{10}' = 10 \times 1.1 = 11 \text{ mm} \]

Final Answer:

The distance of the 10th bright fringe from the central maximum is: \[ \boldsymbol{11 \text{ mm}} \]

Answer 2

Step 1: Understand the given data.
We are performing a double-slit interference experiment with the following data:
Wavelength of light \( \lambda_1 = 600 \, \text{nm} \)
Position of the 10th bright fringe from the central maximum \( y_1 = 10 \, \text{mm} \)
We need to find the new position \( y_2 \) of the same 10th bright fringe when the wavelength is changed to \( \lambda_2 = 660 \, \text{nm} \).

Step 2: Recall the fringe position formula.
For the \( n \)-th bright fringe in an interference pattern:
\[ y = \frac{n \lambda D}{d} \] where \( D \) is the distance between the screen and the slits, and \( d \) is the distance between the slits.
Since \( D \) and \( d \) remain constant, we can write:
\[ \frac{y_2}{y_1} = \frac{\lambda_2}{\lambda_1} \]

Step 3: Substitute the given values.
\[ y_2 = y_1 \times \frac{\lambda_2}{\lambda_1} = 10 \times \frac{660}{600} = 10 \times 1.1 = 11 \, \text{mm} \]

Step 4: Conclusion.
When the wavelength increases from 600 nm to 660 nm, the entire fringe pattern expands, and the 10th bright fringe shifts outward.

Final Answer:
\[ \boxed{11 \, \text{mm}} \]