The angles of incidence and refraction of a monochromatic ray of light of wavelength `lamda` at an air-glass interface are i and r, respectively. A parallel beam of light with a small spread `delta lamda` in wavelength about a mean wavelength `lamda` is refracted at the same air-glass interface. The refractive index `mu` of glass depends on the wavelength `lamda` as `mu(lamda) = a + b//lamda2` where a and b are constants. Then the angular spread in the angle of refraction of the beam is

Options

A. `|(sin i)/(lamda^(3)cos r)delta lamda|`
B. `|(2b)/(lamda^(3))delta lamda|`
C. `|(2b tan r)/(a lamda^(3) +b lamda)delta lamda|`
D. `|(2b(a+b//lamda^(2))sin i)/(lamda^(3))delta lamda|`

Answer

To solve the problem, we need to find the angular spread in the angle of refraction (denoted as \( \Delta r \)) when a parallel beam of light with a small spread \( \Delta \lambda \) in wavelength is refracted at the air-glass interface.

### Step-by-Step Solution:

1. **Understanding the Refractive Index**:
The refractive index \( \mu \) of glass is given as:
\[ \mu(\lambda) = a + \frac{b}{\lambda^2} \]
where \( a \) and \( b \) are constants.

2. **Applying Snell's Law**:
According to Snell's Law:
\[ n_1 \sin(i) = n_2 \sin(r) \]
For air, \( n_1 = 1 \), so:
\[ \sin(i) = \mu \sin(r) \]

3. **Substituting the Refractive Index**:
Substitute \( \mu \) into Snell's Law:
\[ \sin(i) = \left(a + \frac{b}{\lambda^2}\right) \sin(r) \]

4. **Differentiating with Respect to Wavelength**:
Differentiate both sides with respect to \( \lambda \):
\[ 0 = \frac{d\mu}{d\lambda} \sin(r) + \mu \cos(r) \frac{dr}{d\lambda} \]

5. **Finding \( \frac{d\mu}{d\lambda} \)**:
Differentiate \( \mu \):
\[ \frac{d\mu}{d\lambda} = \frac{d}{d\lambda}\left(a + \frac{b}{\lambda^2}\right) = -\frac{2b}{\lambda^3} \]

6. **Substituting \( \frac{d\mu}{d\lambda} \) into the Equation**:
Substitute \( \frac{d\mu}{d\lambda} \) into the differentiated Snell's Law:
\[ 0 = -\frac{2b}{\lambda^3} \sin(r) + \left(a + \frac{b}{\lambda^2}\right) \cos(r) \frac{dr}{d\lambda} \]

7. **Rearranging for \( \frac{dr}{d\lambda} \)**:
Rearranging gives:
\[ \frac{dr}{d\lambda} = \frac{2b \sin(r)}{\lambda^3 \cos(r)} \cdot \frac{1}{a + \frac{b}{\lambda^2}} \]

8. **Finding the Angular Spread**:
The angular spread \( \Delta r \) can be expressed as:
\[ \Delta r = \frac{dr}{d\lambda} \Delta \lambda \]
Substituting the expression for \( \frac{dr}{d\lambda} \):
\[ \Delta r = \frac{2b \sin(r)}{\lambda^3 \cos(r)} \cdot \frac{1}{a + \frac{b}{\lambda^2}} \Delta \lambda \]

9. **Final Expression**:
Thus, the angular spread in the angle of refraction is:
\[ \Delta r = \frac{2b \tan(r)}{a \lambda^2 + b} \Delta \lambda \]

### Final Answer:
The angular spread in the angle of refraction of the beam is:
\[ \Delta r = \frac{2b \tan(r)}{a \lambda^2 + b} \Delta \lambda \]