Practicing Success
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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 |
1.22 1.33 1.51 1.83 |
1.33 |
$2μt\, \cos r =(2n+1)\frac{λ}{2}$ or $μ=\frac{(2n+1)}{2t\, \cos r}\frac{λ}{2}=\frac{(2n+1)×5320×10^{-10}}{2×5×10^{-5}×10^{-2}×1}$ $⇒μ=1.33$ |
