Light is incident from glass $(\mu=1.50)$ to water $(\mu=1.33).$ Find the range of the angle of deviation for which there are two angles of incidence.
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$\mu_\text{g}=1.5=\frac{3}{2}; \ \mu_\text{w}=1.33=\frac{4}{3}$For two angles of incidence,
$\Rightarrow\sin\theta_\text{C}=\frac{8}{9}$
$\Rightarrow\sin^{-1}\Big(\frac{8}{9}\Big)=62.73^{\circ}$
$\therefore$ Angle of deviation $= 90^{\circ} - \theta_\text{C} = 90^{\circ}=\sin^{-1}\Big(\frac{8}{9}\Big)=\cos^{-1}\Big(\frac{8}{9}\Big)=37.27^{\circ}$
Here, if the angle of incidence is increased beyond critical angle, total internal reflection occurs and deviation decreases. So, the range of deviation is 0 to $\cos^{-1}\Big(\frac{8}{9}\Big).$
- When light passes straight through normal,
- When light is incident at critical angle,
$\Rightarrow\sin\theta_\text{C}=\frac{8}{9}$
$\Rightarrow\sin^{-1}\Big(\frac{8}{9}\Big)=62.73^{\circ}$
$\therefore$ Angle of deviation $= 90^{\circ} - \theta_\text{C} = 90^{\circ}=\sin^{-1}\Big(\frac{8}{9}\Big)=\cos^{-1}\Big(\frac{8}{9}\Big)=37.27^{\circ}$
Here, if the angle of incidence is increased beyond critical angle, total internal reflection occurs and deviation decreases. So, the range of deviation is 0 to $\cos^{-1}\Big(\frac{8}{9}\Big).$
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