2. A monochromatic light with wavelength 540.0 nm strikes a pair of narrow slits. An interference...

2. A monochromatic light with wavelength 540.0 nm strikes a pair of narrow slits. An interference pattern is produced on a screen kept 4.00 m away. The first dark fringe is formed at a distance 5.40 mm away from the center. (a) What is the separation between the two slits? [5]

(b) What is the distance on the screen from the center of the interference pattern to the 3rd minimum (m = 2)? [5]

(c) What is the shortest distance from the center where the intensity will reduce to 1/4 of the maximum intensity?

Expert Solution

For this We have to consider the general case of Young Double slit experiment (YDSE) , in which we take slit separation = d , Screen distance = D , and the linear distance of any fringe from central fringe as Y.
Hope so I have solved your doubt. Thanks and regards!

genius_generous answered 2 years ago

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