The correct options are
A 429 nm
B 600 nm
The light of wavelength λ is strongly reflected if 2μd=(n+12)λ ----- (i)
where n is a non negitive integr.
Here 2μd=2×1.50×0.5×10−6 m=1.5×10−6 m
Putting λ=400 nm in Eq. (i), we get 1.5×106m=(n+12)(400×10−9m)⇒n=3.25
putting λ=700 nm in Eq. (i),
we get
1.5×10−6 m=(n+12(700×10−9 m
⟹n=1.66
Thus, within 400 nm to 700 nm, the integer n can take the values 2 and 3. Putting these values of n in E.q (i), the wavelengths become λ=4μd2n+1=600 nm and 429 nm.
Thus, light of wavelengths 429 nm and 600 nm are strongly reflected.