In Y.D.S.E. , distance between both the slits is 2 mm. When a light of frequency 5.5 × 10 14 H z is used, the fringes width is 1 mm, calculate the distance second bright fringe and central fringe.

Answer

To solve the problem step by step, let's break down the information given and apply the relevant formulas.

### Given Data:
- Distance between the slits (d) = 2 mm = 2 × 10^-3 m
- Frequency of light (f) = 5.5 × 10^14 Hz
- Fringe width (β) = 1 mm = 1 × 10^-3 m

### Step 1: Understand the Concept of Fringe Width
The fringe width (β) in Young's Double Slit Experiment (Y.D.S.E) is given by the formula:
\[ \beta = \frac{\lambda D}{d} \]
where:
- \( \lambda \) = wavelength of light
- \( D \) = distance from the slits to the screen
- \( d \) = distance between the slits

### Step 2: Calculate the Wavelength (λ)
First, we need to find the wavelength (λ) using the frequency (f) and the speed of light (c). The speed of light is approximately \( c = 3 × 10^8 \) m/s. The relationship between wavelength, frequency, and speed of light is given by:
\[ \lambda = \frac{c}{f} \]
Substituting the values:
\[ \lambda = \frac{3 × 10^8 \text{ m/s}}{5.5 × 10^{14} \text{ Hz}} \approx 5.45 × 10^{-7} \text{ m} \]

### Step 3: Calculate the Distance of the Second Bright Fringe
The distance of the nth bright fringe from the central maximum is given by:
\[ y_n = n \beta \]
where \( n \) is the order of the fringe. For the second bright fringe (n = 2):
\[ y_2 = 2 \beta = 2 × (1 × 10^{-3} \text{ m}) = 2 × 10^{-3} \text{ m} = 2 \text{ mm} \]

### Step 4: Calculate the Distance of the Central Fringe
The central fringe (n = 0) is located at:
\[ y_0 = 0 \beta = 0 \]
Thus, the distance of the central fringe from the central maximum is 0 mm.

### Final Answers:
- Distance of the second bright fringe from the central fringe = 2 mm
- Distance of the central fringe from the central maximum = 0 mm