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 \[\sin \,C=\frac{1}{_{2}{{\mu }_{1}}}\] \[\sin C=\tan \theta \] \[\Rightarrow \] \[C={{\sin }^{-1}}(tan\theta )\]You need to login to perform this action.
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