Lights of wavelengths `lambda_(1)=4500 Å, lambda_(2)=6000 Å` are sent through a double slit arrangement simultaneously. Then

Options

A. no interference pattern will be formed
B. the third order bright fringe of `lambda_(1)` will coincide with the fourth order bright fringe of `lambda_(2)`
C. the third order bright fringe of `lambda_(2)` will coincide with the fourth order bright fringe of `lambda_(1)`
D. the fringes of wavelength `lambda_(1)` will be wider than the fringes of wavelength `lambda_(2)`

Answer

To solve the problem, we need to find the conditions under which the bright fringes of two different wavelengths coincide in a double slit arrangement.

### Step-by-Step Solution:

1. **Identify the Wavelengths**:
We are given two wavelengths:
- \( \lambda_1 = 4500 \, \text{Å} \)
- \( \lambda_2 = 6000 \, \text{Å} \)

2. **Understand the Condition for Bright Fringes**:
The condition for bright fringes in a double slit experiment is given by the formula:
\[ d \sin \theta = m \lambda \]
where \( d \) is the distance between the slits, \( \theta \) is the angle of the fringe, \( m \) is the order of the fringe, and \( \lambda \) is the wavelength.

3. **Set Up the Equation for Coinciding Bright Fringes**:
For the \( p \)-th bright fringe of wavelength \( \lambda_1 \) and the \( q \)-th bright fringe of wavelength \( \lambda_2 \), we can write:
\[ p \lambda_1 = q \lambda_2 \]

4. **Express the Ratio of Orders**:
Rearranging the equation gives us:
\[ \frac{p}{q} = \frac{\lambda_2}{\lambda_1} \]

5. **Substitute the Values**:
Substitute the values of \( \lambda_1 \) and \( \lambda_2 \):
\[ \frac{p}{q} = \frac{6000 \, \text{Å}}{4500 \, \text{Å}} = \frac{6000}{4500} = \frac{4}{3} \]

6. **Cross Multiply to Find Orders**:
From the ratio \( \frac{p}{q} = \frac{4}{3} \), we can express it as:
\[ 3p = 4q \]
This means that for every 4th order bright fringe of \( \lambda_2 \), there is a corresponding 3rd order bright fringe of \( \lambda_1 \).

7. **Conclusion**:
Therefore, the 4th order bright fringe of \( \lambda_1 \) coincides with the 3rd order bright fringe of \( \lambda_2 \).

### Final Answer:
The third order bright fringe of \( \lambda_2 \) will coincide with the fourth order bright fringe of \( \lambda_1 \).