-->

Wave Optics

Question
CBSEENPH12039395

A monochromatic light of wavelength 500 nm is incident normally on a single slit of width 0.2 mm to produce a diffraction pattern. Find the angular width of the central maximum obtained on the screen.

Estimate the number of fringes obtained in Young's double slit experiment with fringe width 0.5 mm, which can be accommodated within the region of the total angular spread of the central maximum due to the single slit.

Solution

Angular width of central maximum
straight omega space equals space fraction numerator 2 straight lambda over denominator straight a end fraction
space equals space fraction numerator 2 space straight x 500 space straight x 10 to the power of negative 9 end exponent over denominator 0.2 space straight x 10 to the power of negative 3 end exponent end fraction
space equals space 5 space straight x 10 to the power of negative 3 end exponent space radian
straight beta space equals space λD over straight d
Linear width of central maxima in the diffraction pattern
straight omega apostrophe space equals space fraction numerator 2 λD over denominator straight a end fraction
Let ‘n’ be the number of interference fringes which can be accommodated in the central maxima
therefore space straight n space space straight x space straight beta space equals space straight omega apostrophe
straight n space equals space fraction numerator 2 λD over denominator straight a end fraction space straight x straight d over λD
straight n space equals space fraction numerator 2 straight d over denominator straight a end fraction