Question

In a Young's double slit experiment, the slits are separated by 0.3 mm and the screen is 1.5 m away from the plane of slits. Distance between fourth bright fringes on both sides of central bright fringe is 2.4 cm. The frequency of light used is ________ \(\times 10^{14}\) Hz.

Hint:

Be careful with the wording "distance between... on both sides". It means from \(-4^{th}\) fringe to \(+4^{th}\) fringe, which spans a total width of 8 fringe widths.

Answer 1

Correct Answer: 5

Step 1: Understanding the Concept:
The distance of the \(n^{th}\) bright fringe from the central maximum is \( y_n = \frac{n \lambda D}{d} \). The distance between fringes on either side is twice this value.
Step 2: Key Formula or Approach:
1. Distance between \(n^{th}\) bright fringes on both sides: \( \Delta y = 2 y_n = \frac{2 n \lambda D}{d} \).
2. Frequency: \( f = c/\lambda \).
Step 3: Detailed Explanation:
Given: \( d = 0.3 \text{ mm} = 3 \times 10^{-4} \text{ m} \), \( D = 1.5 \text{ m} \), \( n = 4 \), \( \Delta y = 2.4 \text{ cm} = 2.4 \times 10^{-2} \text{ m} \).
Calculate wavelength \(\lambda\):
\[ 2.4 \times 10^{-2} = \frac{2 \times 4 \times \lambda \times 1.5}{3 \times 10^{-4}} \]
\[ 2.4 \times 10^{-2} = \frac{12 \lambda}{3 \times 10^{-4}} = 4 \times 10^4 \lambda \]
\[ \lambda = \frac{2.4 \times 10^{-2}}{4 \times 10^4} = 0.6 \times 10^{-6} \text{ m} = 600 \text{ nm} \]
Calculate frequency \(f\):
\[ f = \frac{c}{\lambda} = \frac{3 \times 10^8}{600 \times 10^{-9}} = 0.005 \times 10^{17} = 5 \times 10^{14} \text{ Hz} \]
Step 4: Final Answer:
The frequency is \( 5 \times 10^{14} \) Hz. The numerical value is 5.