In an interference experiment, third bright fringe is obtained at a point on the screen with a light of 700 nm . What should be the wavelength of the light source in order to obtain 5th bright fringe at the same point

Options

A. 630 nm
B. 500 nm
C. 420 nm
D. 750 nm

Answer

To solve the problem, we need to find the wavelength of light (λ2) that will produce the 5th bright fringe at the same position where the 3rd bright fringe of light with a wavelength of 700 nm is observed.

### Step-by-Step Solution:

1. **Understanding the Condition for Bright Fringes**:
The position of the bright fringes in an interference pattern is given by the formula:
\[ y = n \cdot \lambda \]
where \(y\) is the position of the fringe, \(n\) is the order of the fringe (an integer), and \(\lambda\) is the wavelength of the light.

2. **Setting Up the Equation**:
For the first light source with wavelength λ1 (700 nm), the position of the 3rd bright fringe (n1 = 3) can be expressed as:
\[ y = n_1 \cdot \lambda_1 = 3 \cdot 700 \text{ nm} \]
For the second light source with wavelength λ2, we want the position of the 5th bright fringe (n2 = 5) to be the same:
\[ y = n_2 \cdot \lambda_2 = 5 \cdot \lambda_2 \]

3. **Equating the Two Expressions**:
Since both expressions for \(y\) must be equal, we can set them equal to each other:
\[ 3 \cdot 700 \text{ nm} = 5 \cdot \lambda_2 \]

4. **Solving for λ2**:
Rearranging the equation to solve for λ2 gives:
\[ \lambda_2 = \frac{3 \cdot 700 \text{ nm}}{5} \]
\[ \lambda_2 = \frac{2100 \text{ nm}}{5} = 420 \text{ nm} \]

5. **Final Answer**:
Therefore, the wavelength of the light source needed to obtain the 5th bright fringe at the same point is:
\[ \lambda_2 = 420 \text{ nm} \]