The index of refraction of a glass plate is 1.48 at `theta_(1) = 30.^@C` and varies linearly with temperature with a coefficient of `2.5 xx 10^(-5).^@C^(-1)`. The coefficient of linear expansion of the glass is `5 xx 10^(-9)^@C^(-1)`. At 3`0.^@C`, the length of the glass plate is 3 cm. This plate is placed in front of one of the slit n Young's double-slit experiment. If the plate is being heated so that it temperature increases at a rate of `5^(@)C^(-1)` min, the light sources has wavelength `lambda = 589 nm` and the glass plate initially ia at `theta = 30^@C`. The number of fringes that shift on the screen in each minute is nearly (use approximation)

Options

A. 1
B. 11
C. 110
D. `1.1 xx 10^(3)`

Explanation

To solve the problem step by step, we will follow the outlined process to determine the number of fringes that shift on the screen in one minute due to the heating of the glass plate.

### Step 1: Identify Given Values
- Index of refraction at \( \theta_1 = 30^\circ C \) is \( \mu_0 = 1.48 \)
- Coefficient of refraction change with temperature: \( \alpha_1 = 2.5 \times 10^{-5} \, ^\circ C^{-1} \)
- Coefficient of linear expansion of glass: \( \alpha_2 = 5 \times 10^{-9} \, ^\circ C^{-1} \)
- Length of the glass plate at \( 30^\circ C \): \( L_0 = 3 \, \text{cm} = 3 \times 10^{-2} \, \text{m} \)
- Rate of temperature increase: \( \frac{d\theta}{dt} = 5 \, ^\circ C \, \text{min}^{-1} \)
- Wavelength of light: \( \lambda_0 = 589 \, \text{nm} = 589 \times 10^{-9} \, \text{m} \)

### Step 2: Calculate Change in Index of Refraction
The index of refraction after one minute can be expressed as:
\[ \mu_t = \mu_0 \left(1 + \alpha_1 \Delta \theta\right) \]
Where \( \Delta \theta = 5 \, ^\circ C \) (the temperature increase in one minute).

Substituting the values:
\[ \mu_t = 1.48 \left(1 + 2.5 \times 10^{-5} \times 5\right) \]
Calculating \( \alpha_1 \Delta \theta \):
\[ \alpha_1 \Delta \theta = 2.5 \times 10^{-5} \times 5 = 1.25 \times 10^{-4} \]
Thus,
\[ \mu_t = 1.48 \left(1 + 1.25 \times 10^{-4}\right) \approx 1.48 \times 1.000125 \approx 1.480185 \]

### Step 3: Calculate Path Difference
The path difference \( \Delta x \) due to the change in index of refraction is given by:
\[ \Delta x = (\mu_t - \mu_0) L_0 \]
Substituting the values:
\[ \Delta x = (1.480185 - 1.48) \times (3 \times 10^{-2}) = 0.000185 \times 0.03 = 5.55 \times 10^{-6} \, \text{m} \]

### Step 4: Calculate Number of Fringes Shifted
The number of fringes shifted \( n \) is given by:
\[ n = \frac{\Delta x}{\lambda_0} \]
Substituting the values:
\[ n = \frac{5.55 \times 10^{-6}}{589 \times 10^{-9}} \approx 9.42 \]
Since we are looking for the number of fringes that shift, we round this to the nearest whole number:
\[ n \approx 9 \]

### Step 5: Final Result
The number of fringes that shift on the screen in one minute is approximately **9**.