In Young’s double slit experiment, the distance between slits and screen is 2 m and distance between slits is 0.25 mm. A light of wavelength 800 nm is used to find fringes on the screen. If screen moves with a speed of 5 m/s, then first maxima will move with a speed of

Options

A. 16 mm/s
B. 8 mm/s
C. 20 m/s
D. 50 mm/s

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the given values
- Distance between the slits and the screen (D) = 2 m
- Distance between the slits (d) = 0.25 mm = 0.25 × 10^-3 m
- Wavelength of light (λ) = 800 nm = 800 × 10^-9 m
- Speed of the screen (Vs) = 5 m/s

### Step 2: Calculate the fringe width (β)
The fringe width (β) in Young's double slit experiment is given by the formula:
\[ \beta = \frac{\lambda D}{d} \]
Substituting the values:
\[ \beta = \frac{800 \times 10^{-9} \text{ m} \times 2 \text{ m}}{0.25 \times 10^{-3} \text{ m}} \]

### Step 3: Simplify the calculation
Calculating the numerator:
\[ 800 \times 10^{-9} \times 2 = 1600 \times 10^{-9} \text{ m} \]
Now, substituting into the formula for β:
\[ \beta = \frac{1600 \times 10^{-9}}{0.25 \times 10^{-3}} = \frac{1600 \times 10^{-9}}{0.25 \times 10^{-3}} = \frac{1600}{0.25} \times 10^{-9 + 3} \]
Calculating \( \frac{1600}{0.25} \):
\[ \frac{1600}{0.25} = 6400 \]
Thus,
\[ \beta = 6400 \times 10^{-6} = 6.4 \times 10^{-3} \text{ m} \]

### Step 4: Calculate the speed of the first maxima (Vmax)
The speed of the first maxima can be calculated using the formula:
\[ V_{max} = \frac{d\beta}{dt} = \frac{\lambda V_s}{d} \]
Substituting the values:
\[ V_{max} = \frac{800 \times 10^{-9} \text{ m} \times 5 \text{ m/s}}{0.25 \times 10^{-3} \text{ m}} \]

### Step 5: Simplify the calculation for Vmax
Calculating the numerator:
\[ 800 \times 10^{-9} \times 5 = 4000 \times 10^{-9} \text{ m/s} \]
Now substituting into the formula for Vmax:
\[ V_{max} = \frac{4000 \times 10^{-9}}{0.25 \times 10^{-3}} = \frac{4000 \times 10^{-9}}{0.25 \times 10^{-3}} = \frac{4000}{0.25} \times 10^{-9 + 3} \]
Calculating \( \frac{4000}{0.25} \):
\[ \frac{4000}{0.25} = 16000 \]
Thus,
\[ V_{max} = 16000 \times 10^{-6} = 16 \times 10^{-3} \text{ m/s} = 16 \text{ mm/s} \]

### Final Answer
The speed of the first maxima is:
\[ \boxed{16 \text{ mm/s}} \]