For a single slit of width \'\'a\'\' the first minimum of the interference pattern of a monochromatic light of wavelength e occurs at an angle of `(lambda)/(a)`. At the same angle of `(lambda)/(a)`, we get a maximum for two narrow slits separated by distance \'\'a\'\'. Explain.

The difference in the patterns observed in the single-slit and double-slit experiments is due to the difference in the interference conditions for these two setups.

1. Single-Slit Diffraction: In a single-slit setup, the light waves from different parts of the slit interfere with each other. The condition for destructive interference (which leads to a minimum in the intensity pattern) is that the path difference between the waves from the two edges of the slit is an integer multiple of the wavelength. This condition is met at an angle of (lambda)/(a), where lambda is the wavelength of the light and a is the width of the slit.

2. Double-Slit Interference: In a double-slit setup, the light waves from the two slits interfere with each other. The condition for constructive interference (which leads to a maximum in the intensity pattern) is that the path difference between the waves from the two slits is an integer multiple of the wavelength. This condition is met at an angle of (lambda)/(a), where lambda is the wavelength of the light and a is the separation between the slits.

So, the same angle (lambda)/(a) corresponds to a minimum in the single-slit diffraction pattern and a maximum in the double-slit interference pattern because of the different interference conditions in these two setups.