Visible light of wavelength 550nm falls on a single slit and produced its second diffraction minimum at an angle of 45∘ relative to the incident direction of the light. What is the width of the slit (inμm)?
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Solution
Given: λ=550nm θ=45∘forn=2
We know that the condition for nth minimum is given by
dsinθ=nλ
Where, d = width of slit
⇒dsin45∘=2×550nm
⇒d×1√2=2×550nm
=2×550×√2×10−9m
=1555.63×10−9m=1.56×10−6m
=1.56μm
Why this question? Caution:θnD=nλd can only be applied whenθis very small but in this particular questionθ=45∘ Which is large.