The user wants me to solve a physics problem about light passing through a prism and find the rate of change of the angle of emergence with respect to the refractive index.

1. Identify the given information and relevant equations:

• Angle of incidence, $i_1 = 60^\circ$.
• The prism is equilateral, so the angle of the prism, $A = 60^\circ$.
• The refractive index of the prism is $n$.
• The angle of emergence is $\theta(n)$.
• We are given a specific case: when $n = \sqrt{3}$, the angle of emergence is $\theta = 60^\circ$.
• We need to find $m = \frac{d\theta}{dn}$ at this specific condition.

The behavior of light passing through the prism is described by Snell's law at the two faces and the geometry of the prism:

• First face (incidence): $1 \cdot \sin i_1 = n \sin r_1$
• Second face (emergence): $n \sin r_2 = 1 \cdot \sin \theta$
• Prism geometry: $A = r_1 + r_2$

where $r_1$ is the angle of refraction at the first face and $r_2$ is the angle of incidence at the second face.

2. Analyze the specific case ($n = \sqrt{3}$): Let's find the values of $r_1$ and $r_2$ for this case.

• Using Snell's law at the first face with $i_1 = 60^\circ$ and $n = \sqrt{3}$: $\sin 60^\circ = \sqrt{3} \sin r_1$ $\frac{\sqrt{3}}{2} = \sqrt{3} \sin r_1$ $\sin r_1 = \frac{1}{2} \implies r_1 = 30^\circ$

• Using the prism geometry equation with $A = 60^\circ$: $r_2 = A - r_1 = 60^\circ - 30^\circ = 30^\circ$

• We can verify this with Snell's law at the second face, given $\theta = 60^\circ$: $n \sin r_2 = \sqrt{3} \sin 30^\circ = \sqrt{3} \cdot \frac{1}{2} = \frac{\sqrt{3}}{2}$ $\sin \theta = \sin 60^\circ = \frac{\sqrt{3}}{2}$ The values are consistent. This condition, where $i_1=\theta$ and $r_1=r_2$, corresponds to the minimum deviation condition.

3. Find the derivative $\frac{d\theta}{dn}$: We will differentiate the three prism equations with respect to $n$. Remember that $r_1$, $r_2$, and $\theta$ are all functions of $n$, while $i_1$ and $A$ are constants.

• Differentiating $\sin i_1 = n \sin r_1$: $0 = \frac{d}{dn}(n \sin r_1) = 1 \cdot \sin r_1 + n \cos r_1 \frac{dr_1}{dn}$ $\frac{dr_1}{dn} = -\frac{\sin r_1}{n \cos r_1} = -\frac{\tan r_1}{n}$

• Differentiating $r_1 + r_2 = A$: $\frac{dr_1}{dn} + \frac{dr_2}{dn} = 0 \implies \frac{dr_2}{dn} = -\frac{dr_1}{dn} = \frac{\tan r_1}{n}$

• Differentiating $n \sin r_2 = \sin \theta$: $1 \cdot \sin r_2 + n \cos r_2 \frac{dr_2}{dn} = \cos \theta \frac{d\theta}{dn}$

Now, substitute the expression for $\frac{dr_2}{dn}$ into the last equation: $\sin r_2 + n \cos r_2 \left(\frac{\tan r_1}{n}\right) = \cos \theta \frac{d\theta}{dn}$ $\sin r_2 + \cos r_2 \tan r_1 = \cos \theta \frac{d\theta}{dn}$ $\sin r_2 + \cos r_2 \frac{\sin r_1}{\cos r_1} = \cos \theta \frac{d\theta}{dn}$ $\frac{\sin r_2 \cos r_1 + \cos r_2 \sin r_1}{\cos r_1} = \cos \theta \frac{d\theta}{dn}$

Using the trigonometric identity $\sin(x+y) = \sin x \cos y + \cos x \sin y$: $\frac{\sin(r_1 + r_2)}{\cos r_1} = \cos \theta \frac{d\theta}{dn}$

Since $r_1 + r_2 = A$: $\frac{\sin A}{\cos r_1} = \cos \theta \frac{d\theta}{dn}$

Solving for $\frac{d\theta}{dn}$: $\frac{d\theta}{dn} = \frac{\sin A}{\cos r_1 \cos \theta}$

4. Calculate the value of $m$: We need to evaluate this derivative at the given condition: $n = \sqrt{3}$. From step 2, we have the corresponding angles:

• $A = 60^\circ$
• $r_1 = 30^\circ$
• $\theta = 60^\circ$

Substitute these values into the expression for the derivative: $m = \frac{d\theta}{dn} \bigg|_{n=\sqrt{3}} = \frac{\sin 60^\circ}{\cos 30^\circ \cos 60^\circ}$ $m = \frac{\frac{\sqrt{3}}{2}}{\left(\frac{\sqrt{3}}{2}\right) \left(\frac{1}{2}\right)}$ $m = \frac{1}{\frac{1}{2}} = 2$

The value of m is 2.