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656.


Problem 46.11 (RHK)


Monochromatic light with wavelength 538 nm falls on a slit of width . The distance from the slit to a screen is 3.48 m. We consider a point on the screen 1.13 cm from the central maximum. (a) We have to calculate ; (b) ; (c) and the ratio of the intensity at this point to the intensity at the central maximum.


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 (a)

The distance D of the screen from the slit is 3.48 m. That is

.

The angle subtended by the point P at a distance 1.13 cm from the central maximum on the screen with the line joining the midpoint of the slit with the screen normally will be approximately given by

(b)

If the slit width is denoted by and the wavelength of the light used for producing the diffraction pattern is denoted by , the variable ,which determines the intensity of diffraction, is defined as

.

The wavelength of the monochromatic light used in the experiment is . Therefore, the value of at the point P on the screen will be

(c)

The intensity at a point P subtending an angle relative to the intensity of the central maximum is given by the relation