Question
Half angular width of the central maxima in the fraunhaufer diffraction due to slit width $\frac{{1200}}{{\sqrt 2 }}\ \mu m $ is $45^o$. Then wavelength of the light is......$\mu m $
✓
Answer
$\lambda = d\sin \theta $
$ = \frac{{1200}}{{\sqrt 2 }} \times \frac{1}{{\sqrt 2 }} = 600$
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