Question

A physics teacher wants to demonstrate interference with the help of double slit experiment using a laser beam of 633 nm wavelength. Since the hall is large enough, interference pattern is formed on the wall 5.0 m from the slits. For clear and comfortable view by all the students they want the fringe width 5 mm.

Answer 1

Calculations for Young's Double Slit Experiment

Provided Data:

• Wavelength, \( \lambda = 633 \, \text{nm} = 633 \times 10^{-9} \, \text{m} \)
• Screen distance, \( D = 5.0 \, \text{m} \)
• Fringe width, \( \beta = 5 \, \text{mm} = 5 \times 10^{-3} \, \text{m} \)

(I) Determination of Slit Separation (\( d \))

The fringe width in Young’s Double Slit Experiment is governed by the formula:

\[ \beta = \frac{\lambda D}{d} \]

To find the slit separation \( d \), the formula is rearranged as:

\[ d = \frac{\lambda D}{\beta} \]

Upon substituting the given values:

\[ d = \frac{633 \times 10^{-9} \times 5}{5 \times 10^{-3}} = \frac{3165 \times 10^{-9}}{5 \times 10^{-3}} = 0.000633 \, \text{m} = 0.633 \, \text{mm} \]

Calculated slit separation: \( d = 0.633 \, \text{mm} \)

(II) Calculation of the First Minimum's Distance from the Central Maximum

The distance of the first minimum from the central maximum is given by the formula:

\[ y = \frac{\lambda D}{2d} \]

Using the previously determined values:

\[ y = \frac{633 \times 10^{-9} \times 5}{2 \times 0.000633} = \frac{3165 \times 10^{-9}}{0.001266} = 2.5 \times 10^{-3} \, \text{m} = 2.5 \, \text{mm} \]

Calculated distance of the first minimum from the central maximum: \( y = 2.5 \, \text{mm} \)

Summary of Results:

• Slit separation, \( d = 0.633 \, \text{mm} \)
• Distance of the first minimum from the central maximum, \( y = 2.5 \, \text{mm} \)