Question

Monochromatic light of wavelength 466 nm from a distant source passes through a slit that is 0.0330 mm wide. In the resulting diffraction pattern, the intensity at the center of the central maximum (? = 0?) is 1.06×10^?4 W/m2 . What is the intensity at a point on the screen that corresponds to ? = 1.20?.

Answer 1

The intensity at a point on the screen that makes an angle with the center is given by

where

in radians

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Here

Intensity at the central maximum = 1.06×10^-4 W/m²

Slit width = 0.033 mm = 0.033 x 10^-3 m = 3.3 x 10^-5 m

466 nm = 466 x 10^-9 m

1.2^o

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= [ ( x 3.3 x 10^-5 ) / 466 x 10^-9 ] x = 4.659 radians

So we have, = =

=> = = 0.045937 x 1.06×10^-4 W/m² = 4.8693 x 10^-6 W/m²

4.87 x 10^-6 W/m² ( up to 3 significant figures )

• The intensity at a point on the screen that corresponds to is 4.87 x 10^-6 W/m²