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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

User SandyBr
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