in fraunhofer diffraction pattern slit width is 0.2mm and screen is at 2m away from the lens if wavelength of light used is 5000A then the distance between the first minimum on either side of the central maximum is
Answer (1)
19th Apr, 2020
Hey,
We know that,
the distance between the first 2 minimum of either side of the central maximum is equal to 2 lambda D/ d
where,
it is given that,
d=0.2mm
D= 2 m
lambda=5000A
is,
= (2*5*10^3*10^-10*2)/(2*10^-1*10^-3)
=10*(10^-7/10^-4)
=10^-2 m
which is the required answer.
I hope this helps.
All the best!
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