A long narrow horizontal slit is placed 1 mm above a horizontal plane mirror. The interference between the light coming directly from the slit and that after reflection is seen on a screen 1.0 m away from the slit, Find the fringe-width if the light used has a wavelength of 700 nm.

Explanation

To solve the problem, we need to find the fringe width (β) in an interference pattern created by light from a slit and its reflection from a mirror. Here’s a step-by-step solution:

### Step 1: Understand the Setup
We have a slit (S) that is 1 mm above a horizontal plane mirror. The light from the slit travels directly to the screen and also reflects off the mirror. The distance from the slit to the screen (D) is given as 1.0 m.

### Step 2: Determine the Effective Distance Between the Slit and Its Image
The image of the slit (S') formed by the mirror will be 1 mm below the mirror. Therefore, the distance between the slit and its image is:
- Distance from the slit to the mirror = 1 mm = 1 × 10^(-3) m
- Distance from the mirror to the image of the slit = 1 mm = 1 × 10^(-3) m
- Total distance between the slit and its image (d) = 1 mm + 1 mm = 2 mm = 2 × 10^(-3) m

### Step 3: Use the Formula for Fringe Width
The formula for fringe width (β) in an interference pattern is given by:
\[ \beta = \frac{\lambda D}{d} \]
where:
- λ = wavelength of light = 700 nm = 700 × 10^(-9) m
- D = distance from the slit to the screen = 1 m
- d = distance between the slit and its image = 2 mm = 2 × 10^(-3) m

### Step 4: Substitute the Values into the Formula
Now we can substitute the values into the formula:
\[ \beta = \frac{(700 \times 10^{-9} \text{ m}) \times (1 \text{ m})}{(2 \times 10^{-3} \text{ m})} \]

### Step 5: Calculate the Fringe Width
Calculating this gives:
\[ \beta = \frac{700 \times 10^{-9}}{2 \times 10^{-3}} = \frac{700}{2} \times 10^{-6} = 350 \times 10^{-6} \text{ m} = 0.35 \text{ mm} \]

### Final Answer
Thus, the fringe width (β) is **0.35 mm**.

---