A monochromatic beam of light is incident at 60 degrees on one face of an equilateral prism of refra

Q: Why is the angle of the prism $A = 60^{\circ}$?

The problem states it is an equilateral prism. By definition, all three internal angles of an equilateral triangle are $60^{\circ}$.

Q: Why do we differentiate with respect to $n$?

The question asks for the value of $m$, which is defined as the derivative of the emergent angle $\theta$ with respect to the refractive index $n$, i.e., $\frac{d\theta}{dn}$. This requires us to find the rate of change of $\theta$ as $n$ varies.

Q: How did we get $r_1 + r_2 = 60^{\circ}$?

In any prism, the sum of the angles of refraction at the two interfaces ($r_1$ and $r_2$) is always equal to the angle of the prism $A$. Since $A = 60^{\circ}$, $r_1 + r_2 = 60^{\circ}$.

Q: Why is the derivative of $\sin 60^{\circ}$ zero?

The angle of incidence at the first face is fixed at $60^{\circ}$. It does not change when the refractive index $n$ changes. Since it's a constant, its derivative is zero.

Q: Can we just substitute $n = \sqrt{3}$ before differentiating?

No! If you substitute the constant value of $n$ before differentiating, you will just get $0 = 0$. You must differentiate the general equations first to find the relationship between the rates of change, and *then* substitute the specific values for the given instant.

Question

A monochromatic beam of light is incident at $60^{\circ}$ on one face of an equilateral prism of refractive index $n$ and emerges from the opposite face making an angle $	heta(n)$ with the normal (see figure). For $n = \sqrt{3}$ the value of $	heta$ is $60^{\circ}$ and $\frac{d	heta}{dn} = m$. The value of $m$ is

Sigma Percentile · Accepted Answer

The problem asks us to find the rate of change of the angle of emergence $	heta$ with respect to the refractive index $n$ of an equilateral prism, evaluated at $n = \sqrt{3}$. This is a beautiful application of Snell's law combined with implicit differentiation.

### Analyzing the Setup

Let's visualize the path of the monochromatic light beam through the equilateral prism. The beam hits the first face at an angle of incidence $i_1 = 60^{\circ}$, refracts inside at an angle $r_1$, and emerges from the second face at an angle of emergence $i_2 = 	heta$.

For an equilateral prism, the prism angle $A$ is $60^{\circ}$. We know from the geometry of prisms that the sum of the internal angles of refraction equals the prism angle:
$$r_1 + r_2 = A = 60^{\circ}$$

### The Master Equation

To find how $	heta$ changes with the refractive index $n$, we need to set up the equations for refraction. Applying Snell's law at the first surface:
$$\sin 60^{\circ} = n \sin r_1$$

Applying Snell's law at the second surface:
$$n \sin r_2 = \sin 	heta$$

Since $r_2 = 60^{\circ} - r_1$, we can rewrite the second equation as:
$$n \sin (60^{\circ} - r_1) = \sin 	heta$$

Now, we differentiate the first equation with respect to $n$. The derivative of $\sin 60^{\circ}$ is zero because the angle of incidence is constant. Using the product rule on the right side, we get:
$$0 = \sin r_1 + n \cos r_1 \frac{dr_1}{dn}$$

Rearranging this gives us the rate of change of the first refraction angle:
$$\frac{dr_1}{dn} = -\frac{\sin r_1}{n \cos r_1} = -\frac{	an r_1}{n}$$

Next, we differentiate the second equation with respect to $n$:
$$\cos 	heta \frac{d	heta}{dn} = \sin (60^{\circ} - r_1) + n \cos (60^{\circ} - r_1) \left(-\frac{dr_1}{dn}ight)$$

We can substitute the expression for $\frac{dr_1}{dn}$ into this equation. The $n$ cancels out, and the minus signs become plus:
$$\cos 	heta \frac{d	heta}{dn} = \sin (60^{\circ} - r_1) - n \cos (60^{\circ} - r_1) \left(-\frac{	an r_1}{n}ight)$$
$$\cos 	heta \frac{d	heta}{dn} = \sin (60^{\circ} - r_1) + \cos (60^{\circ} - r_1) 	an r_1$$

### Final Calculation

We are given that $n = \sqrt{3}$. Let's find the specific angles for this case. From the first equation:
$$\sin 60^{\circ} = \sqrt{3} \sin r_1$$
$$\frac{\sqrt{3}}{2} = \sqrt{3} \sin r_1 \Rightarrow \sin r_1 = \frac{1}{2}$$

So, $r_1 = 30^{\circ}$. Consequently, $r_2 = 60^{\circ} - 30^{\circ} = 30^{\circ}$.

Now, let's find the emergent angle $	heta$:
$$\sin 	heta = \sqrt{3} \sin 30^{\circ} = \frac{\sqrt{3}}{2} \Rightarrow 	heta = 60^{\circ}$$

Finally, let's substitute these angles into our differentiated equation:
$$\cos 60^{\circ} \frac{d	heta}{dn} = \sin (60^{\circ} - 30^{\circ}) + \cos (60^{\circ} - 30^{\circ}) 	an 30^{\circ}$$
$$\frac{1}{2} \frac{d	heta}{dn} = \sin 30^{\circ} + \cos 30^{\circ} 	an 30^{\circ}$$
$$\frac{1}{2} \frac{d	heta}{dn} = \frac{1}{2} + \frac{\sqrt{3}}{2} 	imes \frac{1}{\sqrt{3}}$$
$$\frac{1}{2} \frac{d	heta}{dn} = \frac{1}{2} + \frac{1}{2} = 1$$

Therefore, $\frac{d	heta}{dn} = 2$. The value of $m$ is **2**.

Enter Numerical Value:

Visualized Solution

Visualizing the Prism

Applying Snell's Law

Differentiating the First Equation

Differentiating the Second Equation

Substituting the Derivative

Evaluating Angles for $n = 3$

Final Calculation

The Sigma Insight: Refraction and Dispersion through Prism

Analyzing the Setup

The Master Equation

Final Calculation

Similar Questions

A monochromatic light is incident at a certain angle on an equilateral triangular prism and suffers minimum deviation. If the refractive index of the material of the prism is $3$ , then the angle of incidence is

A ray of light undergoes deviation of $3 0^{\circ}$ when incident on an equilateral prism of refractive index $2$ . The angle made by the ray inside the prism with the base of the prism is ______

A ray of light passing through a prism ( $μ = 3$ ) suffers minimum deviation. It is found that the angle of incidence is double the angle of refraction within the prism. Then, the angle of prism is ……… (in degrees).

Monochromatic light is incident on a glass prism of angle $A$ . If the refractive index of the material of the prism is $μ$ , a ray incident at an angle $θ$ , on the face $A B$ would get transmitted through the face $A C$ of the prism provided

An isosceles prism of angle $12 0^{\circ}$ has a refractive index $1.44$ . Two parallel rays of monochromatic light enter the prism parallel to each other in air as shown. The rays emerging from the opposite face

Enter Numerical Value:

Visualized Solution

Visualizing the Prism

Applying Snell's Law

Differentiating the First Equation

Differentiating the Second Equation

Substituting the Derivative

Evaluating Angles for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n=3​

Final Calculation

The Sigma Insight: Refraction and Dispersion through Prism

Analyzing the Setup

The Master Equation

Final Calculation

Similar Questions

Evaluating Angles for $n = 3$