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 find the wavelength of light used in a double-slit experiment, we can follow these steps:

### Step 1: Understand the given data
- Distance between the first maxima and fifth minima, \( \Delta y = 7 \, \text{mm} \)
- Slit separation, \( d = 0.15 \, \text{mm} \)
- Distance from slits to screen, \( D = 50 \, \text{cm} \)

### Step 2: Convert units
- Convert \( \Delta y \) to meters:
\[ \Delta y = 7 \, \text{mm} = 7 \times 10^{-3} \, \text{m} \]
- Convert \( d \) to meters:
\[ d = 0.15 \, \text{mm} = 0.15 \times 10^{-3} \, \text{m} \]
- Convert \( D \) to meters:
\[ D = 50 \, \text{cm} = 0.50 \, \text{m} \]

### Step 3: Relate maxima and minima positions to fringe width
- The position of the first maxima (m=1) is given by:
\[ y_1 = \frac{1}{2} \frac{\lambda D}{d} \]
- The position of the fifth minima (m=5) is given by:
\[ y_5 = \frac{5}{2} \frac{\lambda D}{d} \]
- The distance between the first maxima and fifth minima is:
\[ \Delta y = y_5 - y_1 = \left(\frac{5}{2} \frac{\lambda D}{d}\right) - \left(\frac{1}{2} \frac{\lambda D}{d}\right) = \frac{4}{2} \frac{\lambda D}{d} = 2 \frac{\lambda D}{d} \]

### Step 4: Set up the equation
- From the distance between the first maxima and fifth minima, we have:
\[ 2 \frac{\lambda D}{d} = 7 \times 10^{-3} \, \text{m} \]

### Step 5: Solve for wavelength \( \lambda \)
- Rearranging gives:
\[ \lambda = \frac{7 \times 10^{-3} \cdot d}{2D} \]
- Substitute the values:
\[ \lambda = \frac{7 \times 10^{-3} \cdot (0.15 \times 10^{-3})}{2 \cdot 0.50} \]
- Calculate:
\[ \lambda = \frac{7 \times 0.15 \times 10^{-6}}{1} = 1.05 \times 10^{-6} \, \text{m} = 1050 \, \text{nm} \]

### Step 6: Final result
- The wavelength of the light used is \( \lambda = 1050 \, \text{nm} \).