Question 276: Ray Optics

Question

A ray of light suffers minimum deviation when incident on a prism having angle of the prism equal to $60^{\circ}$. The refractive index of the prism material is $\sqrt{2}$. The angle of incidence (in degrees) is__________

Accepted Answer

To find the angle of incidence when a ray of light experiences minimum deviation in a prism, we use the formula for the refractive index $\mu$:
$ \mu = \frac{\sin \left(\frac{A + \delta_m}{2}\right)}{\sin \left(\frac{A}{2}\right)} $
Given:
The angle of the prism, $A = 60^{\circ}$
The refractive index of the prism material, $\mu = \sqrt{2}$
Minimum deviation, $\delta_m = 30^{\circ}$
At minimum deviation, the angle of incidence $i$ equals the angle of emergence $e$. Therefore, the relation between the angle of minimum deviation and the angle of the prism is represented as:
$ \delta_m = 2i - A $
Solving for $i$:
$ \delta_m = 2i - A \implies 30^{\circ} = 2i - 60^{\circ} \implies 2i = 90^{\circ} \implies i = 45^{\circ} $
Thus, the angle of incidence is $45^{\circ}$.

Ray Optics

A ray of light suffers minimum deviation when incident on a prism having angle of the prism equal to 60^{\circ}. The refractive index of the prism material is \sqrt{2}. The angle of incidence (in degrees) is__________

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