In a single-slit experiment, the slit width is 150 times the wavelength of the light.
What is the width (in mm) of the central maximum on a screen 2.6 m behind the slit?
I have tried:
y=[(1+1/2)(lambda)(2.6m)] / (150lambda)
to bring me to
y=[(1.5)(2.6)] \ (150)
giving me a y value in mm of 26
Answers: 2
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