Ray Optics

JEE Main level, single correct, medium

60 of 295

Question 60 of 295

A ray of light passes from a denser medium to a rarer medium at an angle of incidence i.i. The reflected and refracted rays make an angle of 90∘90^{\circ} with each other. The angle of reflection and refraction are respectively rr and r′.{r}^{'}. The critical angle is given by,

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Answer: (D) sin⁡−1(tan⁡r){\sin }^{-1}(\tan r)

r+r′+90∘=180∘⇒r′=90−r=90−ir+{r}^{'}+90^{\circ}=180^{\circ}\Rightarrow {r}^{'}=90-r=90-i
n1sin⁡i=n2sin⁡r′=n2sin⁡(90−i){n}_{1}\sin i={n}_{2}\sin {r}^{'}={n}_{2}\sin (90-i)
n1sin⁡i=n2cos⁡i⇒tan⁡i=n2n1{n}_{1}\sin i={n}_{2}\cos i\Rightarrow \tan i=\dfrac{{n}_{2}}{{n}_{1}}
Now sin⁡C=n2n1=tan⁡i\sin C=\dfrac{{n}_{2}}{{n}_{1}}=\tan i
⇒C=sin⁡−1(tan⁡i)=sin⁡−1(tan⁡r)\Rightarrow C={\sin }^{-1}(\tan i)={\sin }^{-1}(\tan r)

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