Concept:In single slit diffraction, secondary maxima are formed approximately midway between adjacent minima. For the second maximum, the path difference condition is used with a small-angle approximation.
Explanation:For the second secondary maximum, the condition is:
asinθ=25λSince the diffraction angle is small, we can write:
sinθ≈tanθ=DySubstituting this into the condition gives:
a⋅Dy=25λRearranging for
λ, we get:
λ=5D2ayNow convert the given values to SI units:
a=0.66 mm=0.66×10−3 my=2.8 mm=2.8×10−3 mD=1.4 mSubstitute these values:
λ=5×1.42×0.66×10−3×2.8×10−3λ=5.28×10−7 mSince
1A˚=10−10m, we have:
λ=5280A˚Thus, the wavelength of light used is
5280A˚.
Answer:Option C:
5280A˚