Monochromatic
light of wavelength 441 nm falls on a narrow slit.
On a screen 2.16 m
away, the
distance between the second minimum and the central maximum
is 1.62 cm. (a) We
have to calculate the angle of diffraction
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Solution: Click For PDF Version Let the slit width be .
For a monochromatic light of wavelength
the condition for diffraction minima is
Data of the problem are: The distance between the second minimum and central maximum is
And, the distance of the screen from the slit is
The angle
from the line joining the centre of the slit with the central
maximum
The angle
We note that
or
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of the second minimum.
(b) We have to find the width of the slit.
will be
.
