If the distance between the first maxima and fifth minima of a double-slit pattern is 7 mm and the slits are separated by 0.15 mm with the screen 50 cm from the slits, then wavelength of the light used is

Options

A. 600 nm
B. 525 nm
C. 467 nm
D. 420 nm

Explanation

To solve the problem, we need to find the wavelength of the light used in a double-slit experiment given the distance between the first maxima and the fifth minima, the slit separation, and the distance to the screen.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Distance between the first maxima and the fifth minima, \( Y = 7 \, \text{mm} = 7 \times 10^{-3} \, \text{m} \)
- Slit separation, \( d = 0.15 \, \text{mm} = 0.15 \times 10^{-3} \, \text{m} \)
- Distance from the slits to the screen, \( D = 50 \, \text{cm} = 0.5 \, \text{m} \)

2. **Formulas for Maxima and Minima:**
- The distance of the \( n \)-th maxima from the central maximum is given by:
\[ Y_n = \frac{n D \lambda}{d} \]
- The distance of the \( n \)-th minima from the central maximum is given by:
\[ Y_n = \frac{(2n - 1) D \lambda}{2d} \]

3. **Calculate the Distances:**
- For the first maxima (\( n = 1 \)):
\[ Y_1 = \frac{1 \cdot D \cdot \lambda}{d} = \frac{D \lambda}{d} \]
- For the fifth minima (\( n = 5 \)):
\[ Y_5 = \frac{(2 \cdot 5 - 1) D \lambda}{2d} = \frac{9 D \lambda}{2d} \]

4. **Set Up the Equation:**
- The distance between the first maxima and the fifth minima is:
\[ Y_5 - Y_1 = 7 \times 10^{-3} \, \text{m} \]
- Substituting the expressions for \( Y_5 \) and \( Y_1 \):
\[ \frac{9 D \lambda}{2d} - \frac{D \lambda}{d} = 7 \times 10^{-3} \]
- Simplifying the left side:
\[ \frac{9 D \lambda}{2d} - \frac{2 D \lambda}{2d} = \frac{(9 - 2) D \lambda}{2d} = \frac{7 D \lambda}{2d} \]
- Thus, we have:
\[ \frac{7 D \lambda}{2d} = 7 \times 10^{-3} \]

5. **Solve for Wavelength \( \lambda \):**
- Canceling \( 7 \) from both sides:
\[ \frac{D \lambda}{2d} = 10^{-3} \]
- Rearranging gives:
\[ \lambda = \frac{2d \cdot 10^{-3}}{D} \]
- Substituting the values:
\[ \lambda = \frac{2 \cdot (0.15 \times 10^{-3}) \cdot (10^{-3})}{0.5} \]
- Calculating:
\[ \lambda = \frac{0.3 \times 10^{-6}}{0.5} = 0.6 \times 10^{-6} \, \text{m} = 600 \, \text{nm} \]

### Final Answer:
The wavelength of the light used is \( \lambda = 600 \, \text{nm} \).