Ray Optics

JEE Main level, single correct, medium

67 of 295

Question 67 of 295

A ray of light is incident from a denser to a rarer medium. The critical angle for total internal reflection is θiC{\theta }_{\text{iC}} and Brewster's angle of incidence is θiB{\theta }_{\text{iB}}, such that sinθiCsinθiB=η=1.28\dfrac{\text{sin}{\theta }_{\text{iC}}}{\text{sin}{\theta }_{\text{iB}}}=\eta =1\text{.}28. The relative refractive index of the two media is

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Answer: (D) 0.8\text{0.8}

μ1:{\mu }_{1}: refractive index of rarer medium

μ2:{\mu }_{2}: retractive index of denser medium

sinθC=μ1μ2{sin\theta }_{C}=\dfrac{{\mu }_{1}}{{\mu }_{2}}

tanθB=μ1μ2{tan\theta }_{B}=\dfrac{{\mu }_{1}}{{\mu }_{2}}

sin⁡θC=tan⁡θB\sin {\theta }_{C}=\tan {\theta }_{B}

sin⁡θCsin⁡θB=tan⁡θBsin⁡θB=1cos⁡θB=1.28\dfrac{\sin {\theta }_{C}}{\sin {\theta }_{B}}=\dfrac{\tan {\theta }_{B}}{\sin {\theta }_{B}}=\dfrac{1}{\cos {\theta }_{B}}=1.28

tan⁡θB=μ1μ2=1−(11.28)2(11.28)\tan {\theta }_{B}=\dfrac{{\mu }_{1}}{{\mu }_{2}}=\dfrac{\sqrt{1-{(\dfrac{1}{1.28})}^{2}}}{(\dfrac{1}{1.28})}

=0.8=0.8

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