In Young’s slit experiment, the separation of the first to fifth fringes is 2.5 mm when the wavelength used is 620 nm. The distance from the slits to the screen is 80 cm. Calculate the separation of two slits.

Solution
Separation of first and 5^th bright fringe (y) = 2.5 mm = 2.5 x 10^-3 m
Here, n = 4
So, [latex]\beta = \frac{y}{n}[/latex]

= [latex]\frac{2.5\times 10^{-3}}{4} = 6.25\times 10^{-4}[/latex]   m
Wavelength of light used ([latex]\lambda[/latex]) = 620 nm
            = [latex]620\times 10^{-9}[/latex] m = 6.2 x 10^-7 m
Distance from the slit to the screen (D) = 80 cm = 0.8 m
Separation of the slits (d) =?
We know that,
[latex]\beta = \frac{\lambda D}{d}[/latex]
Or, d = [latex]\frac{\lambda D}{\beta}[/latex]
            = [latex]\frac{(6.2\times 10^{-7})\times 0.8}{(6.25\times 10^{-4})}[/latex] = 7.936 x 10^-4 m.

In a Young’s double slit experiment, the separation of four bright fringes is 2.5 mm. The wavelength of light used is 6.2 x 10⁻⁵ cm and the distance from the slits to the screen is 80 cm. Calculate the separation of slits.

Solution
Given,
Separation of four bright fringes (y) = [latex]3\beta[/latex] = 2.5 mm = [latex]2.5\times 10^{-3}[/latex] m
Wavelength of light used [latex](\lambda)[/latex] = [latex]6.2\times 10^{-5}[/latex] cm = [latex]6.2\times 10^{-7}[/latex] m
Distance from the slits to the screen (D) = 80 cm = 0.8 m
Separation of slits (d) = ?

We know that,

[latex]y = 3\beta = \frac{3\lambda D}{d}[/latex]

Or, [latex]d = \frac{3\lambda D}{y}[/latex]

[latex]= \frac{3\times(6.2\times 10^{-7})\times 0.8}{(2.5\times 10^{-3})}[/latex]

[latex]= 5.952\times 10^{-4}[/latex] m.

The distance between two coherent sources in Young’s double slit experiment is 0.3 mm and the interference pattern is observed on a screen 60 cm from the sources. If the wavelength of light used is 6 x 10⁻⁷ m, calculate the fringe width of the interference pattern.

Solution
Given,
Distance between two coherent sources (d) = 0.3 mm = [latex]0.3\times 10^{-3}[/latex] m
Distance of screen from the sources (D) = 60 cm = 0.6 m
Wavelength of light used [latex](\lambda)[/latex] = [latex]6\times 10^{-7}[/latex] m
Fringe width [latex](\beta)[/latex] = ?

We know that,