Ray Optics

JEE Main level, single correct, medium

266 of 295

Question 266 of 295

In an experiment for determination of refractive index of glass of a prism by ii v/s δ\delta plot, it was found that a ray incident at angle 35o{35}^{o} , suffers a deviation of 40o{40}^{o} and that it emerges at angle 79o{79}^{o} . In that case which of the following is closest to the maximum possible value of the refractive index?

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Answer: (C) 1.5

i=35o,δ=40o,e=79oi={35}^{o},\delta ={40}^{o},e={79}^{o}

δ=i+e−A\delta =i+e-A

40o=35o+79o−A{40}^{o}={35}^{o}+{79}^{o}-A

A=74oA={74}^{o}

And r1+r2=A=74o{r}_{1}+{r}_{2}=A={74}^{o}

Since δmin<40o{\delta }_{min}\lt {40}^{o}

μ<sin⁡⁡(74+402)sin⁡⁡37\mu \lt \dfrac{\sin ⁡\left(\dfrac{74+40}{2}\right)}{\sin ⁡37}

μmax=1.44{\mu }_{max}=1.44

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