A beam of polychromatic light passes through a thin prism of prism angle 6 degrees. The refractive i

Q: Why do we use the formula $\delta = A(n - 1)$ instead of the general prism formula?

The problem states it is a "thin prism" with an angle of $6^\circ$. For prism angles less than about $10^\circ$, the small angle approximation $\sin(\theta) \approx \theta$ is valid, which simplifies the general deviation formula to $\delta = A(n - 1)$.

Q: What does the equation $n(\lambda) = \alpha\lambda + \frac{\beta}{\lambda^2}$ represent?

This is a modified empirical relationship similar to Cauchy's equation, which describes how the refractive index of a material changes with the wavelength of light. It mathematically models the phenomenon of dispersion.

Q: Why do we set the derivative $\frac{d\delta}{d\lambda}$ to zero?

In calculus, the local minimum or maximum of a smooth function occurs where its rate of change (the derivative) is zero. Since we want the "smallest" angle of minimum deviation, we are looking for the minimum of the function $\delta(\lambda)$.

Q: How do we know setting the derivative to zero gives a minimum and not a maximum?

You can verify this by taking the second derivative $\frac{d^2\delta}{d\lambda^2} = A\left(\frac{6\beta}{\lambda^4}\right)$. Since $A$, $\beta$, and $\lambda$ are all positive, the second derivative is positive, confirming that the critical point is indeed a minimum.

Q: Is $\lambda_{\text{min}}$ the minimum wavelength of light?

No, $\lambda_{\text{min}}$ is just a variable name chosen in the question to represent the *specific* wavelength at which the deviation $D_m$ reaches its minimum value. It does not mean the shortest possible wavelength.

Question

A beam of polychromatic light passes through a thin prism of prism angle $6^\circ$. The refractive index of the material of the prism varies with wavelength ($\lambda$) as $n(\lambda) = \alpha\lambda + \frac{\beta}{\lambda^2}$, where $\alpha = 3 \,\mu	ext{m}^{-1}$ and $\beta = 0.096 \,\mu	ext{m}^2$. If $\lambda_{	ext{min}}$ is the wavelength at which the angle of minimum deviation $D_m$ is smallest, then the correct value of $D_m$ at $\lambda_{	ext{min}}$ is

Options:

(A) $6.4^\circ$

(B) $4.8^\circ$

(C) $3.2^\circ$

(D) $2.4^\circ$

Sigma Percentile · Accepted Answer

### The Calculus of Colors: Finding the Minimum Deviation

Imagine a beam of white light hitting a glass prism. We all know the beautiful result: a rainbow of colors spreading out. This happens because different colors (which correspond to different wavelengths) bend by different amounts. This phenomenon is called **dispersion**.

In this problem, we are dealing with a very specific, thin prism with an angle of just $6^\circ$. For such thin prisms, the angle of deviation $\delta$ is practically independent of the angle of incidence and is given by a beautifully simple formula:

$$\delta = A(n - 1)$$

Here, $A$ is the prism angle and $n$ is the refractive index of the material. But there's a catch! The refractive index $n$ isn't a single fixed number. It dances to the tune of the wavelength $\lambda$, following a mathematical relationship given to us as:

$$n(\lambda) = \alpha\lambda + \frac{\beta}{\lambda^2}$$

### The Master Equation

To understand how the deviation changes with the color of light, we need to merge our two equations. By substituting the expression for $n$ into our deviation formula, we get a master equation that links deviation directly to wavelength:

$$\delta(\lambda) = A\left(\alpha\lambda + \frac{\beta}{\lambda^2} - 1ight)$$

Now, the question throws a curveball: it asks for the *smallest* angle of minimum deviation, denoted as $D_m$. Whenever physics asks for a "minimum" or "maximum," it's a secret invitation to use calculus.

### The Power of the Derivative

To find the minimum of our function $\delta(\lambda)$, we must take its derivative with respect to $\lambda$ and set it to zero. This is the mathematical equivalent of finding the bottom of a valley on a graph.

$$\frac{d\delta}{d\lambda} = A\left(\alpha - \frac{2\beta}{\lambda^3}ight) = 0$$

Since the prism angle $A$ is $6^\circ$ (and definitely not zero), the expression inside the parentheses must be zero. This gives us the critical condition for minimum deviation:

$$\alpha = \frac{2\beta}{\lambda^3}$$

Rearranging this to solve for the wavelength cubed, we get:

$$\lambda^3 = \frac{2\beta}{\alpha}$$

### Crunching the Numbers

Now, it's time to bring in the numbers. We are given $\alpha = 3 \,\mu	ext{m}^{-1}$ and $\beta = 0.096 \,\mu	ext{m}^2$. Let's substitute these into our condition:

$$\lambda^3 = \frac{2 	imes 0.096}{3} = \frac{0.192}{3} = 0.064$$

Taking the cube root of $0.064$ is surprisingly neat. We find that the specific wavelength $\lambda_{	ext{min}}$ that produces the smallest deviation is:

$$\lambda_{	ext{min}} = 0.4 \,\mu	ext{m}$$

### The Final Calculation

We have the wavelength, but the question asks for the actual angle of deviation $D_m$. To find this, we must first determine the refractive index $n$ at this specific wavelength. Let's plug $\lambda = 0.4$ back into our original equation for $n$:

$$n = 3(0.4) + \frac{0.096}{(0.4)^2}$$

$$n = 1.2 + \frac{0.096}{0.16}$$

$$n = 1.2 + 0.6 = 1.8$$

So, at the wavelength that minimizes deviation, the prism material has a refractive index of $1.8$.

Finally, we substitute this value back into our thin prism formula to find the grand answer:

$$D_m = A(n - 1)$$

$$D_m = 6^\circ (1.8 - 1)$$

$$D_m = 6^\circ 	imes 0.8 = \mathbf{4.8^\circ}$$

And there we have it! By combining the physics of optics with the optimization power of calculus, we've elegantly arrived at the smallest angle of minimum deviation.

Select Answer:

Visualized Solution

\text{Thin Prism Deviation}

\text{Deviation as a function of } \lambda

\text{Condition for Minimum Deviation}

\text{Solving for } \lambda

\text{Calculating } \lambda_{\text{min}}

\text{Refractive Index at } \lambda_{\text{min}}

\text{Calculating } n

\text{Final Minimum Deviation}

The Sigma Insight: Refraction and Dispersion through Prism

The Calculus of Colors

The Master Equation

The Power of the Derivative

Crunching the Numbers

The Final Calculation

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