Question

The critical angle for the material of a prism is 45° and its refracting angle is 30°.A monochromatic ray goes out perpendicular to the surface of emergence from the prism .Then the angle of incidence on the prism will be

Answer is 45°

Answer 1

Answer on Question #71182, Physics / Optics

Question. The critical angle for the material of a prism is $45{}^{\circ}$ and its refracting angle is $30{}^{\circ}$. A monochromatic ray goes out perpendicular to the surface of emergence from the prism. Then the angle of incidence on the prism will be

Given. $\alpha = 30{}^{\circ}; \theta_{c} = 45{}^{\circ}$

Find. $\beta$

Solution.

Using the equation for the critical angle

$$\theta_{c} = \arcsin \left(\frac{n_{2}}{n_{1}}\right)$$

we get

$$45{}^{\circ} = \arcsin \left(\frac{1}{n_{1}}\right) \rightarrow n_{1} = \frac{1}{\sin 45{}^{\circ}} = \sqrt{2}.$$

From the figure

$$\gamma = \alpha.$$

So

$$\frac{\sin \beta}{\sin \gamma} = \frac{n_{1}}{n_{2}} \rightarrow \sin \beta = \frac{n_{1} \cdot \sin \gamma}{n_{2}} \rightarrow$$ $$\beta = \arcsin \left(\frac{n_{1} \cdot \sin \gamma}{n_{2}}\right) = \arcsin \left(\frac{\sqrt{2} \cdot \sin 30{}^{\circ}}{1}\right) = \arcsin \left(\frac{\sqrt{2}}{2}\right) \rightarrow$$ $$\beta = 45{}^{\circ}$$

Answer. $\beta = 45{}^{\circ}$

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