Interference And Diffraction

To find the value of in the given Young's double-slit experiment, we start by analyzing the relationship between slit widths and intensities in the interference pattern. Assume the electric field amplitude at the screen is proportional to the slit width.

Let's denote the amplitude of the wave from the first slit as and from the second slit as , where is a constant proportionality factor.

The resulting amplitude at any point on the screen is given by:

The intensity is proportional to the square of the amplitude:

The maximum intensity occurs when the two waves are in phase:

The minimum intensity occurs when the two waves are out of phase, i.e., their amplitudes subtract:

We are given the ratio of the maximum to minimum intensity as 9:4:

Substitute the expressions for and :

This reduces to:

Taking the square root on both sides:

Cross-multiply to solve for :

Expand and simplify:

Combine like terms:

Finally, solve for :

Therefore, the value of is 5.

Concept: For single slit diffraction, the angular width of the central maximum is where = wavelength of light, = slit width.
Step 1:Convert units
Step 2:Calculate angular width in radians
Step 3:Convert to degrees}
Step 4:Express in required form Thus