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A ray of light is incident at an angle of incidence 'i' on one surface of a thin prism of
small angle 'A'. The ray emerges normally from the opposite surface. If the refractive
index of the material of the prism is ' $\mu$ ', the angle of incidence 'i' is nearly equal to
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small angle 'A'. The ray emerges normally from the opposite surface. If the refractive
index of the material of the prism is ' $\mu$ ', the angle of incidence 'i' is nearly equal to
Solution:
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Verified Answer
The correct answer is:
$\mu A$
$\mu=\frac{\sin \mathrm{i}}{\sin \mathrm{r}}=\frac{\mathrm{i}}{\mathrm{r}_{1}} \quad$ if angles are small.
$A=r_{1}+r_{2}=r_{1}$
since $r_{2}=0$ since the ray emerges normally.
$\therefore \mu=\frac{i}{A} \quad$ or $\quad i=\mu A$
$A=r_{1}+r_{2}=r_{1}$
since $r_{2}=0$ since the ray emerges normally.
$\therefore \mu=\frac{i}{A} \quad$ or $\quad i=\mu A$
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