Question

A Fraunhofer diffraction pattern due to a single slit of width \(0.3\,\text{mm}\) is obtained on a screen placed at a distance of \(3\,\text{m}\) from the slit. The first minima lie at \(5.5\,\text{mm}\) on either side of the central maximum on the screen. The wavelength of light used is

• A. \(6000\,\text{\AA}\)
• B. \(5500\,\text{\AA}\)
• C. \(4500\,\text{\AA}\)
• D. \(5000\,\text{\AA}\)

Hint:

In single slit diffraction, the first minima depends directly on slit width and distance of screen.

Answer 1

The Correct Option is B

Step 1: Use the condition for first minima in single slit diffraction.
For Fraunhofer diffraction, the position of first minima is given by \[ a \sin\theta = \lambda \] For small angles, \( \sin\theta \approx \tan\theta = \frac{y}{D} \).

Step 2: Substitute given values.
\[ \lambda = \frac{a y}{D} \] \[ a = 0.3\,\text{mm} = 3\times10^{-4}\,\text{m}, \quad y = 5.5\,\text{mm} = 5.5\times10^{-3}\,\text{m}, \quad D = 3\,\text{m} \] \[ \lambda = \frac{3\times10^{-4}\times5.5\times10^{-3}}{3} = 5.5\times10^{-7}\,\text{m} \]
Step 3: Convert into angstrom.
\[ \lambda = 5500\,\text{\AA} \]