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.
Exercise
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The refracted ray in the prism is incident on the second face at critical angle ic.
Given, angle of prism, A = 60°
Refracting index of the material = 1.524
Now,$ 60^\circ + 90^\circ - r + 90^\circ - i_c=_180^\circ$
and $\text{r}_1+\text{r}_2=\angle\text{A}$
$\Rightarrow \ \text{r}=60^\circ-\text{i}_\text{c}$
$\sin\text{i}_\text{c}=\frac{1}{\mu}=\frac{1}{1.524}$
$\Rightarrow \ \text{i}_\text{c}=\sin^{-1}\Big(\frac{1}{1.524}\Big)$
$\Rightarrow \ \text{i}_\text{c}\simeq41^\circ$
$\therefore \ \text{r}=60^\circ-41^\circ=19^\circ$
Using Snell's law,
$\sin\text{i}=\sin19^\circ\times1.524$
= 0.4962
$\text{i}=\sin^{-1}(0.4962)$
= 29.75°.
Given, angle of prism, A = 60°
Refracting index of the material = 1.524
Now,$ 60^\circ + 90^\circ - r + 90^\circ - i_c=_180^\circ$
and $\text{r}_1+\text{r}_2=\angle\text{A}$
$\Rightarrow \ \text{r}=60^\circ-\text{i}_\text{c}$
$\sin\text{i}_\text{c}=\frac{1}{\mu}=\frac{1}{1.524}$
$\Rightarrow \ \text{i}_\text{c}=\sin^{-1}\Big(\frac{1}{1.524}\Big)$
$\Rightarrow \ \text{i}_\text{c}\simeq41^\circ$
$\therefore \ \text{r}=60^\circ-41^\circ=19^\circ$
Using Snell's law,
$\sin\text{i}=\sin19^\circ\times1.524$
= 0.4962
$\text{i}=\sin^{-1}(0.4962)$
= 29.75°.
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