A beam of light consisting of two wavelengths `650 nm` and `520 nm` is used to obtain interference fringes in a Young's double slit experiment.
(a) Find the distance of the third bright fringe on the screen from the central maximum for the wavelength `650 nm`.
(b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide? The distance between the slits is `2 mm` and the distance between the plane of the slits and screen is `120 cm`.

Options

A. 1.17 mm
B. 2.52 mm
C. 1.56 mm
D. 3.14 mm

Explanation

To solve the problem, we will break it down into two parts as specified in the question.

### Given Data:
- Wavelength 1, \( \lambda_1 = 650 \, \text{nm} = 650 \times 10^{-9} \, \text{m} \)
- Wavelength 2, \( \lambda_2 = 520 \, \text{nm} = 520 \times 10^{-9} \, \text{m} \)
- Distance between the slits, \( d = 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m} \)
- Distance from slits to screen, \( D = 120 \, \text{cm} = 1.2 \, \text{m} \)

### Part (a): Distance of the Third Bright Fringe for Wavelength 650 nm

The distance of the nth bright fringe from the central maximum in a double-slit experiment is given by the formula:
\[ X_n = \frac{n \lambda D}{d} \]
Where:
- \( n \) is the order of the fringe (in this case, \( n = 3 \))
- \( \lambda \) is the wavelength (for this part, \( \lambda = 650 \, \text{nm} \))
- \( D \) is the distance from the slits to the screen
- \( d \) is the distance between the slits

Substituting the values:
\[ X_3 = \frac{3 \times (650 \times 10^{-9}) \times (1.2)}{2 \times 10^{-3}} \]

Calculating:
1. Calculate the numerator:
\[ 3 \times 650 \times 10^{-9} \times 1.2 = 2.34 \times 10^{-6} \, \text{m} \]
2. Now divide by \( d \):
\[ X_3 = \frac{2.34 \times 10^{-6}}{2 \times 10^{-3}} = 1.17 \times 10^{-3} \, \text{m} = 1.17 \, \text{mm} \]

### Part (b): Least Distance from Central Maximum where Bright Fringes Coincide

To find the least distance from the central maximum where the bright fringes due to both wavelengths coincide, we use the condition:
\[ n_1 \lambda_1 = n_2 \lambda_2 \]
Where:
- \( n_1 \) and \( n_2 \) are the orders of the bright fringes for wavelengths \( \lambda_1 \) and \( \lambda_2 \) respectively.

Rearranging gives:
\[ \frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1} \]

Substituting the values:
\[ \frac{n_1}{n_2} = \frac{520}{650} = \frac{4}{5} \]
This implies:
- \( n_1 = 4 \)
- \( n_2 = 5 \)

Now, we will find the distance of the fourth bright fringe (for \( \lambda_1 \)) from the central maximum:
\[ X_{n_1} = \frac{n_1 \lambda_1 D}{d} \]

Substituting the values:
\[ X_4 = \frac{4 \times (650 \times 10^{-9}) \times (1.2)}{2 \times 10^{-3}} \]

Calculating:
1. Calculate the numerator:
\[ 4 \times 650 \times 10^{-9} \times 1.2 = 3.12 \times 10^{-6} \, \text{m} \]
2. Now divide by \( d \):
\[ X_4 = \frac{3.12 \times 10^{-6}}{2 \times 10^{-3}} = 1.56 \times 10^{-3} \, \text{m} = 1.56 \, \text{mm} \]

### Final Answers:
(a) The distance of the third bright fringe from the central maximum for the wavelength 650 nm is \( 1.17 \, \text{mm} \).

(b) The least distance from the central maximum where the bright fringes due to both wavelengths coincide is \( 1.56 \, \text{mm} \).