Question:medium

The polarising angle for a medium is found to be \( 60^\circ \). The critical angle of the medium is

Show Hint

Remember: \(\tan i_p = \mu\) and \(\sin C = 1/\mu\) . Therefore, a very direct shortcut is \(\sin C = 1 / \tan i_p = \cot i_p\) . For \( i_p = 60^\circ \), \( \cot 60^\circ = 1/\sqrt{3} \)
Updated On: May 6, 2026
  • \( \sin^{-1} \left( \frac{1}{2} \right) \)
  • \( \sin^{-1} \left( \frac{\sqrt{3}}{2} \right) \)
  • \( \sin^{-1} \left( \frac{1}{\sqrt{3}} \right) \)
  • \( \sin^{-1} \left( \frac{1}{4} \right) \)
  • \( \sin^{-1} \left( \frac{2}{\sqrt{3}} \right) \)
Show Solution

The Correct Option is C

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