When light is incident on a soap film of thickness $5 × 10^{–5} cm$, wavelength reflected maximum in the visible region is 5320 Å. Refractive index of the film will be
Answer & explanation
Correct answer: option 2
$2μt\, \cos r =(2n+1)\frac{λ}{2}$
or $μ=\frac{(2n+1)}{2t\, \cos r}\frac{λ}{2}=\frac{(2n+1)×5320×10^{-10}}{2×5×10^{-5}×10^{-2}×1}$
$⇒μ=1.33$