Question

In a Fraunhoffer diffraction, light of wavelength '\(\lambda\)' is incident on slit of width ' d '. The diffraction pattern is observed on a screen placed at a distance ' D '. The linear width of central maximum is equal to two times the width of the slit, then 'D' has value

• A. \(\frac{d^2}{\lambda}\)
• B. \(\frac{d^2}{2\lambda}\)
• C. \(\frac{d^2}{3\lambda}\)
• D. \(\frac{d^2}{4\lambda}\)

Hint:

Central maximum width = \(2\lambda D/d\)

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
In single slit diffraction, the central maximum is bounded by the first minima on either side.
The angular width depends on the wavelength and the slit width. The linear width also depends on the distance to the screen.
Step 2: Key Formula or Approach:
Angular position of first minimum: \(\sin \theta \approx \theta = \frac{\lambda}{d}\).
Angular width of central maximum: \(2\theta = \frac{2\lambda}{d}\).
Linear width of central maximum: \(W = (2\theta) \cdot D = \frac{2\lambda D}{d}\).
Step 3: Detailed Explanation:
According to the problem, the linear width \(W\) is equal to two times the slit width \(d\): \[ W = 2d \] Equate the two expressions for \(W\): \[ \frac{2\lambda D}{d} = 2d \] Cancel the factor 2 from both sides: \[ \frac{\lambda D}{d} = d \] Solve for \(D\): \[ \lambda D = d^2 \implies D = \frac{d^2}{\lambda} \] Step 4: Final Answer:
The value of \(D\) is \(\frac{d^2}{\lambda}\).