At what angle should a ray of light be incident on the face of a prism of refracting angle 60° so that it just suffers total internal reflection at the other face? The refractive index of the material of the prism is 1.524.

Solution

The refracted ray in the prism is incident on the second face at critical angle i_c.

Given, angle of prism, A = 60^oRefracting index of the material = 1.524

Now, 60° + 90° – r + 90° – i_c = 180° [ sum of all angles of $∆$ is 180⁰ ]

and,
$r_{1} + r_{2} = ∠ A$

$\Rightarrow$    $r = 60 ° - i_{c}$

Now,    $\sin i_{c} = \frac{1}{μ} = \frac{1}{1.524}$
$\Rightarrow$ $i_{c} = \sin^{- 1} (\frac{1}{1.524})$
$\Rightarrow$    $i_{c} ≃ 41 °$

$∴$ $r = 60 ° - 41 ° = 19 °$

Using Snell's law,

$\sin i = \sin 19 ° \times 1.524 = 0.4962 i = \sin^{- 1} (0.4962) = 29.75 ° .$

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Wave Optics

At what angle should a ray of light be incident on the face of a prism of refracting angle 60° so that it just suffers total internal reflection at the other face? The refractive index of the material of the prism is 1.524.

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