question_answer

Light from a denser medium \[I\] goes to a rarer medium \[II\]. When angle of incidence is \[\theta \], the reflected and refracted rays are perpendicular to each other. The critical angle is :

A) \[{{\sin }^{-1}}(\cos \theta )\]                   

B) \[{{\sin }^{-1}}(\cot \theta )\]

C) \[{{\sin }^{-1}}(\tan \theta )\]                   

D) \[{{\sin }^{-1}}(1)\]

Correct Answer: C

Solution :

                 From laws of reflection, we know that angle of incidence is equal to angle of reflection Let r the angle of refraction, along line AOB, we have \[\theta +{{90}^{o}}+r={{180}^{o}}\] \[\Rightarrow \]          \[r=90-{{\theta }^{o}}\] Also from Snell?s law \[_{1}{{\mu }_{2}}=\frac{\sin i}{\sin r}\] In this \[case\,i=\theta ,\,r={{90}^{o}}-\theta \] \[_{1}{{\mu }_{2}}=\frac{\sin \theta }{\sin ({{90}^{o}}-\theta )}=\frac{\sin \theta }{\cos \theta }=\tan \theta \] Also from definition of critical angle, we have \[\sin \,C=\frac{1}{_{2}{{\mu }_{1}}}\] \[\sin C=\tan \theta \] \[\Rightarrow \]               \[C={{\sin }^{-1}}(tan\theta )\]