In young's double slit experiment, the screen is kept 1.6m from the slits. The coherent sources are 0.032 cm apart and fringes are observed on the screen. It is found that with a certain monochromatic source of light, the fourth bright fringe is situated at a distance of 1.06 cm from the central fringe. The wavelength of the light used is

Options

A. 530 nm
B. 265 nm
C. 1060 nm
D. 132.5 nm

Answer

To solve the problem, we will use the formula for the position of bright fringes in Young's double slit experiment. The formula is given by:

\[ y_n = \frac{n \lambda D}{d} \]

Where:
- \( y_n \) is the distance of the nth bright fringe from the central maximum,
- \( n \) is the fringe order (in this case, \( n = 4 \)),
- \( \lambda \) is the wavelength of the light,
- \( D \) is the distance from the slits to the screen,
- \( d \) is the distance between the two slits.

### Step-by-step Solution:

1. **Convert the given values to meters:**
- Distance from the screen, \( D = 1.6 \, \text{m} \)
- Distance between slits, \( d = 0.032 \, \text{cm} = 0.032 \times 10^{-2} \, \text{m} = 0.00032 \, \text{m} \)
- Distance of the fourth bright fringe from the central fringe, \( y_4 = 1.06 \, \text{cm} = 1.06 \times 10^{-2} \, \text{m} = 0.0106 \, \text{m} \)

2. **Substitute the known values into the formula:**
\[ y_4 = \frac{n \lambda D}{d} \]
For \( n = 4 \):
\[ 0.0106 = \frac{4 \lambda (1.6)}{0.00032} \]

3. **Rearranging the equation to solve for \( \lambda \):**
\[ 0.0106 = \frac{6.4 \lambda}{0.00032} \]
Multiply both sides by \( 0.00032 \):
\[ 0.0106 \times 0.00032 = 6.4 \lambda \]
\[ 3.392 \times 10^{-6} = 6.4 \lambda \]

4. **Solve for \( \lambda \):**
\[ \lambda = \frac{3.392 \times 10^{-6}}{6.4} \]
\[ \lambda = 0.530 \times 10^{-6} \, \text{m} \]

5. **Convert \( \lambda \) from meters to nanometers:**
\[ \lambda = 0.530 \times 10^{-6} \, \text{m} = 530 \, \text{nm} \]

### Final Answer:
The wavelength of the light used is \( \lambda = 530 \, \text{nm} \).

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