Question 279: Ray Optics

Question

The refractive index of prism is $$\mu=\sqrt{3}$$ and the ratio of the angle of minimum deviation to the angle of prism is one. The value of angle of prism is _________$$^\circ$$.

Accepted Answer

To find the value of the angle of the prism given the refractive index of the prism $$\mu = \sqrt{3}$$ and the ratio of the angle of minimum deviation ($$\delta_m$$) to the angle of the prism ($$A$$) is 1 (i.e., $$\frac{\delta_m}{A} = 1 \Rightarrow \delta_m = A$$), we can use the prism formula relating these variables.
The formula that relates the angle of deviation $$\delta$$, refractive index $$\mu$$, angle of prism $$A$$, and angle of minimum deviation $$\delta_m$$ is given by:
$$\mu = \frac{\sin \left(\frac{\delta_m + A}{2}ight)}{\sin \left(\frac{A}{2}ight)}$$.
Given that $$\delta_m = A$$, our formula becomes:
$$\sqrt{3} = \frac{\sin \left(\frac{A + A}{2}ight)}{\sin \left(\frac{A}{2}ight)}$$
Simplifying this, we have:
$$\sqrt{3} = \frac{\sin(A)}{\sin \left(\frac{A}{2}ight)}$$
Now, recall the trigonometric identity:
$$\sin(2	heta) = 2\sin(	heta)\cos(	heta)$$
By setting $$	heta = \frac{A}{2}$$, we get:
$$\sin(A) = 2\sin\left(\frac{A}{2}ight)\cos\left(\frac{A}{2}ight)$$
Substituting back into our equation:
$$\sqrt{3} = \frac{2\sin\left(\frac{A}{2}ight)\cos\left(\frac{A}{2}ight)}{\sin \left(\frac{A}{2}ight)}$$
$$\sqrt{3} = 2\cos\left(\frac{A}{2}ight)$$
Dividing by 2 and solving for $$\cos\left(\frac{A}{2}ight)$$:
$$\cos\left(\frac{A}{2}ight) = \frac{\sqrt{3}}{2}$$
This corresponds to an angle $$\frac{A}{2}$$ of $$30^\circ$$ since the cosine of 30 degrees is $$\frac{\sqrt{3}}{2}$$. Thus:
$$\frac{A}{2} = 30^\circ$$
So, the angle of the prism $$A$$ is:
$$A = 2 	imes 30^\circ = 60^\circ$$.
Therefore, the value of the angle of the prism is $$60^\circ$$.

Ray Optics

The refractive index of prism is $\mu=\sqrt{3} and the ratio of the angle of minimum deviation to the angle of prism is one. The value of angle of prism is _________^\circ$.

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