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 \(n(\lambda) = a\lambda + \frac{b}{\lambda^2}\), where \(a = 3\,\mu\text{m}^{-1}\) and \(b = 0.096\,\mu\text{m}^2\). If \(\lambda_{\min}\) is the wavelength at which the angle of minimum deviation \(D_m\) is smallest, then the correct value of \(D_m\) at \(\lambda_{\min}\) is:

• A. $6.4^\circ$
• B. $4.8^\circ$
• C. $3.2^\circ$
• D. $2.4^\circ$

Hint:

For a thin prism, deviation is directly proportional to $(n-1)$. Minimizing deviation is mathematically identical to minimizing the refractive index of the material.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The angle of deviation for a thin prism depends on the refractive index $n$. Since $n$ is a function of wavelength $\lambda$, the deviation $D_m$ also depends on $\lambda$. We need to find the minimum value of this deviation.

Step 2: Key Formula or Approach:

• Deviation for a thin prism: $D_m = (n-1)A$

• To find minimum $D_m$, we find the minimum of $n(\lambda)$ by setting $\frac{dn}{d\lambda} = 0$.

Step 3: Detailed Explanation:

• Refractive index function: $n(\lambda) = a\lambda + b\lambda^{-2}$

• Differentiate with respect to $\lambda$: \[ \frac{dn}{d\lambda} = a - \frac{2b}{\lambda^3} \]
• For minimum $n$ (and thus minimum $D_m$), set $\frac{dn}{d\lambda} = 0$: \[ a = \frac{2b}{\lambda^3} \implies \lambda^3 = \frac{2b}{a} = \frac{2 \times 0.096}{3} = 0.064 \] \[ \lambda_{min} = (0.064)^{1/3} = 0.4 \mu\text{m} \]
• Calculate the minimum refractive index $n(\lambda_{min})$: \[ n = 3(0.4) + \frac{0.096}{(0.4)^2} = 1.2 + \frac{0.096}{0.16} = 1.2 + 0.6 = 1.8 \]
• Calculate the corresponding deviation: \[ D_m = (1.8 - 1) \times 6^\circ = 0.8 \times 6^\circ = 4.8^\circ \]

Step 4: Final Answer:
The value of $D_m$ at $\lambda_{min}$ is $4.8^\circ$.