Question

A parallel beam of light of wavelength 650 nm passes through a slit of width 0.6 mm. The diffraction pattern is obtained on a screen kept 60 cm away from the slit. Find the distance between first order minima on both sides of the central maximum.

Answer 1

Single Slit Diffraction: Angular Position of First Minima

Given:

• Wavelength \( \lambda = 650 \, \text{nm} = 650 \times 10^{-9} \, \text{m} \)
• Slit width \( a = 0.6 \, \text{mm} = 0.6 \times 10^{-3} \, \text{m} \)
• Distance to the screen \( D = 60 \, \text{cm} = 0.6 \, \text{m} \)

Formula for Angular Position of the First Minima:

The angular position of the first minima in single slit diffraction is given by:

\[ a \sin \theta = m\lambda, \quad \text{for } m = \pm 1 \]

For small angles, \( \sin \theta \approx \tan \theta = \frac{y}{D} \), so the equation becomes:

\[ y = \frac{\lambda D}{a} \]

Step 1: Substituting the Given Values

Substituting the values \( \lambda = 650 \times 10^{-9} \, \text{m} \), \( D = 0.6 \, \text{m} \), and \( a = 0.6 \times 10^{-3} \, \text{m} \) into the equation:

\[ y = \frac{650 \times 10^{-9} \times 0.6}{0.6 \times 10^{-3}} = \frac{390 \times 10^{-9}}{0.6 \times 10^{-3}} = 0.00065 \, \text{m} = 0.65 \, \text{mm} \]

Step 2: Total Distance Between First Minima on Both Sides

The total distance between the first minima on both sides of the central maximum is twice the value of \( y \):

\[ 2y = 2 \times 0.65 \, \text{mm} = 1.3 \, \text{mm} \]

Final Answer:

• Distance to the first minima from the central maximum: \( y = 0.65 \, \text{mm} \)
• Total distance between first minima on both sides: \( 2y = 1.3 \, \text{mm} \)