A beam of light consisting of two wavelengths 6500Å and 5200Å is used to obtain interference fringes in YDSE. The distance between slits is 2mm and the distance of the screen form slits is 120 cm. What is the least distance from central maximum where the bright due to both wavelengths coincide?

Options

A. 0.156cm
B. 0.312cm
C. 0.078cm
D. 0.468cm

Explanation

To solve the problem of finding the least distance from the central maximum where the bright fringes due to both wavelengths coincide in a Young's Double Slit Experiment (YDSE), we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Data:**
- Wavelengths:
- \( \lambda_1 = 6500 \, \text{Å} = 6500 \times 10^{-10} \, \text{m} \)
- \( \lambda_2 = 5200 \, \text{Å} = 5200 \times 10^{-10} \, \text{m} \)
- Distance between slits: \( d = 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m} \)
- Distance from slits to screen: \( D = 120 \, \text{cm} = 1.2 \, \text{m} \)

2. **Write the Formula for Fringe Position:**
The position of the bright fringe on the screen for each wavelength can be given by:
\[ y_n = \frac{n \lambda D}{d} \]
where \( n \) is the order of the fringe.

3. **Set Up the Condition for Coincidence:**
For the bright fringes of both wavelengths to coincide:
\[ y_{n_1} = y_{n_2} \]
This leads to:
\[ \frac{n_1 \lambda_1 D}{d} = \frac{n_2 \lambda_2 D}{d} \]
Simplifying gives:
\[ n_1 \lambda_1 = n_2 \lambda_2 \]

4. **Express the Ratio of Orders:**
Rearranging the above equation gives:
\[ \frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1} \]
Substituting the values of wavelengths:
\[ \frac{n_1}{n_2} = \frac{5200}{6500} = \frac{4}{5} \]
This means that the 4th bright fringe of \( \lambda_2 \) coincides with the 5th bright fringe of \( \lambda_1 \).

5. **Calculate the Position of Coinciding Fringes:**
We can use either fringe order to calculate the position. Let's use \( n_1 = 5 \):
\[ y_1 = \frac{n_1 \lambda_1 D}{d} = \frac{5 \times 6500 \times 10^{-10} \times 1.2}{2 \times 10^{-3}} \]
Calculating this:
\[ y_1 = \frac{5 \times 6500 \times 10^{-10} \times 1.2}{2 \times 10^{-3}} = \frac{5 \times 6500 \times 1.2}{2} \times 10^{-7} \]
\[ = \frac{39000}{2} \times 10^{-7} = 19500 \times 10^{-7} \, \text{m} = 0.00195 \, \text{m} = 1.95 \, \text{cm} \]

### Final Answer:
The least distance from the central maximum where the bright fringes due to both wavelengths coincide is **1.95 cm**.