A light of wavelength 500 nm is incident on a young's slit. The distance between slits annd screen is D=1.8m and distance between slits is d=0.4 mm. if screen moves with s speed 4m/s, with what speed first maxima will move?

Options

A. a. 5mm/s
B. b.4mm/s
C. c.3mm/s
D. d.2mm/s

Explanation

To solve the problem step by step, we will use the information provided and apply the relevant formulas.

### Step 1: Write down the given data
- Wavelength of light, \( \lambda = 500 \, \text{nm} = 500 \times 10^{-9} \, \text{m} \)
- Distance between the slits and the screen, \( D = 1.8 \, \text{m} \)
- Distance between the slits, \( d = 0.4 \, \text{mm} = 0.4 \times 10^{-3} \, \text{m} \)
- Speed of the screen, \( v = 4 \, \text{m/s} \)

### Step 2: Find the fringe width (beta)
The fringe width \( \beta \) can be calculated using the formula:
\[ \beta = \frac{\lambda D}{d} \]
Substituting the values:
\[ \beta = \frac{500 \times 10^{-9} \times 1.8}{0.4 \times 10^{-3}} \]

### Step 3: Calculate the value of beta
Calculating \( \beta \):
\[ \beta = \frac{500 \times 10^{-9} \times 1.8}{0.4 \times 10^{-3}} = \frac{900 \times 10^{-9}}{0.4 \times 10^{-3}} = \frac{900 \times 10^{-9}}{4 \times 10^{-4}} = 2.25 \times 10^{-3} \, \text{m} = 2.25 \, \text{mm} \]

### Step 4: Differentiate the fringe width with respect to time
To find the speed of the first maxima, we differentiate the fringe width with respect to time:
\[ \frac{d\beta}{dt} = \frac{\lambda}{d} \cdot \frac{dD}{dt} \]
Where \( \frac{dD}{dt} \) is the speed of the screen \( v \).

### Step 5: Substitute the values into the differentiated equation
Substituting the known values:
\[ \frac{d\beta}{dt} = \frac{500 \times 10^{-9}}{0.4 \times 10^{-3}} \cdot 4 \]

### Step 6: Calculate the speed of the first maxima
Calculating \( \frac{d\beta}{dt} \):
\[ \frac{d\beta}{dt} = \frac{500 \times 10^{-9}}{0.4 \times 10^{-3}} \cdot 4 = \frac{500 \times 4}{0.4} \times 10^{-6} = \frac{2000}{0.4} \times 10^{-6} = 5000 \times 10^{-6} = 5 \times 10^{-3} \, \text{m/s} = 5 \, \text{mm/s} \]

### Final Answer
The speed of the first maxima is \( 5 \, \text{mm/s} \).

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