You illuminate a slit with a width of 0.0605 mm with a light of wavelength 719...
You illuminate a slit with a width of 0.0605 mm with a light of wavelength 719 nm and observe the resulting diffraction pattern on a screen that is situated 2.33 m from the slit. What is the width, in centimeters, of the pattern's central maximum?
Homework Answers
Solution)
We know,
The angle to the first minimum
sinθ1min=mλ/D =1*719*10^-9/0.0605*10^-3=0.0118
Now,
Y=Lsinθ =2.33*0.0188=0.0276 m
Now, width of the central maximum is
W=2Y=2*0.0276=0.0553
W=5.53 cm
============
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