Question

In a single slit diffraction experiment, the width of the slit and the wavelength of the light are respectively 5 mm and 500 nm. If the focal length of the lens is 20 cm, then the size of the central bright fringe will be

• A. \( 5 \times 10^{-5} \, {m} \)
• B. \( 3 \times 10^{-5} \, {m} \)
• C. \( 2.5 \times 10^{-5} \, {m} \)
• D. \( 2 \times 10^{-5} \, {m} \)
• E. \( 1 \times 10^{-5} \, {m} \)

Hint:

In single slit diffraction, the angular width of the central fringe is inversely proportional to the slit width. Larger slit widths result in smaller fringe sizes.

Answer 1

The Correct Option is D

Step 1: In a single slit diffraction experiment, the angular width of the central diffraction fringe is given by: \[ \theta = \frac{\lambda}{a} \] where \( \lambda \) is the wavelength of light, and \( a \) is the width of the slit.
Step 2: The linear width of the central bright fringe \( Y \) on the screen can be found using: \[ Y = \theta \times f \] where \( f \) is the focal length of the lens.
Step 3: Substituting the values: \[ \lambda = 500 \, {nm} = 5 \times 10^{-7} \, {m}, \quad a = 5 \, {mm} = 5 \times 10^{-3} \, {m}, \quad f = 20 \, {cm} = 0.2 \, {m}. \] Thus, \[ \theta = \frac{5 \times 10^{-7}}{5 \times 10^{-3}} = 1 \times 10^{-4} \, {radians}. \] Step 4: The linear size of the central fringe is: \[ Y = \theta \times f = (1 \times 10^{-4}) \times 0.2 = 2 \times 10^{-5} \, {m}. \] Thus, the size of the central bright fringe is \( 2 \times 10^{-5} \, {m} \).