Question

White light is passed through a double slit and interference is observed on a screen 1.5 m away. The separation between the slits is 0.3 mm. The first violet and red fringes are formed 2.0 mm and 3.5 mm away from the central white fringes. The difference in wavelengths of red and violet light is _________ nm.

Hint:

Fringe width \(\beta\) is directly proportional to wavelength. Since red has a longer wavelength than violet, it always produces wider fringes and appears further from the center for the same fringe order.

Answer 1

Correct Answer: 300

Step 1: Understanding the Concept:
In Young's Double Slit Experiment (YDSE), the position of the \(n\)-th bright fringe from the center is given by \(y_n = \frac{n\lambda D}{d}\). Since violet and red have different wavelengths, their fringes appear at different positions.
Step 2: Key Formula or Approach:
\[ \lambda = \frac{y \cdot d}{n \cdot D} \]
Where \(y\) is the fringe position, \(d\) is slit separation, \(D\) is screen distance, and \(n=1\) for first fringes.
Step 3: Detailed Explanation:
Given: \(D = 1.5 \text{ m}\), \(d = 0.3 \text{ mm} = 3 \times 10^{-4} \text{ m}\).

1. For Violet light: \(y_v = 2.0 \text{ mm} = 2 \times 10^{-3} \text{ m}\).
\[ \lambda_v = \frac{y_v d}{D} = \frac{(2 \times 10^{-3})(3 \times 10^{-4})}{1.5} = \frac{6 \times 10^{-7}}{1.5} = 4 \times 10^{-7} \text{ m} = 400 \text{ nm} \]

2. For Red light: \(y_r = 3.5 \text{ mm} = 3.5 \times 10^{-3} \text{ m}\).
\[ \lambda_r = \frac{y_r d}{D} = \frac{(3.5 \times 10^{-3})(3 \times 10^{-4})}{1.5} = \frac{10.5 \times 10^{-7}}{1.5} = 7 \times 10^{-7} \text{ m} = 700 \text{ nm} \]

3. Difference in wavelengths:
\[ \Delta \lambda = \lambda_r - \lambda_v = 700 - 400 = 300 \text{ nm} \]
Step 4: Final Answer:
The difference in wavelengths is 300 nm.