J & K CET Medical J & K - CET Medical Solved Paper-2010

  • question_answer
    The Brewster angle for the glass-air interface is \[54.74{}^\circ \]. If a ray of light going from air to glass strikes at an angle of incidence \[45{}^\circ \], then the angle of refraction is (Given, tan \[54.74{}^\circ \]= \[\sqrt{2}\])

    A) \[60{}^\circ \]

    B) \[30{}^\circ \]

    C) \[25{}^\circ \]

    D) \[54.74{}^\circ \]

    Correct Answer: B

    Solution :

    Refractive index, \[\mu =\tan \theta =\sqrt{2}\] Now according to Snells law \[\sqrt{2}=\frac{\sin (54.74)}{\sin r}\] \[\sin r=\frac{0.816}{\sqrt{2}}\] \[sin\text{ }r=0.57\] \[r={{35}^{o}}\simeq {{30}^{o}}\]


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