The fringe width in YDSE is `2.4 xx 10^(-4)m`, when red light of wavelength `6400 Å` is used. By how much will it change, if blue light of wavelength `4000 Å` is used ?

Options

A. `9 xx10^(-4)m`
B. `0.9 xx10^(-4)m`
C. `4.5 xx10^(-4)m`
D. `0.45 xx10^(-4)m`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write down the given quantities.
- Fringe width for red light (β_red) = \(2.4 \times 10^{-4} \, m\)
- Wavelength for red light (λ_red) = \(6400 \, Å = 6400 \times 10^{-10} \, m\)
- Wavelength for blue light (λ_blue) = \(4000 \, Å = 4000 \times 10^{-10} \, m\)

### Step 2: Understand the relationship between fringe width and wavelength.
The fringe width (β) in Young's Double Slit Experiment (YDSE) is given by the formula:
\[ \beta = \frac{D \lambda}{d} \]
where:
- D = distance from the slits to the screen (constant)
- d = distance between the slits (constant)

Since D and d are constants, the fringe width is directly proportional to the wavelength (λ):
\[ \beta \propto \lambda \]

### Step 3: Set up the ratio of fringe widths.
Using the proportionality, we can express the fringe width for blue light (β_blue) in terms of the fringe width for red light (β_red):
\[ \frac{\beta_{red}}{\beta_{blue}} = \frac{\lambda_{red}}{\lambda_{blue}} \]
Rearranging gives us:
\[ \beta_{blue} = \beta_{red} \cdot \frac{\lambda_{blue}}{\lambda_{red}} \]

### Step 4: Substitute the known values into the equation.
Substituting the values we have:
\[ \beta_{blue} = (2.4 \times 10^{-4}) \cdot \frac{4000 \times 10^{-10}}{6400 \times 10^{-10}} \]

### Step 5: Simplify the equation.
The \(10^{-10}\) terms cancel out:
\[ \beta_{blue} = (2.4 \times 10^{-4}) \cdot \frac{4000}{6400} \]
Now simplify \(\frac{4000}{6400}\):
\[ \frac{4000}{6400} = \frac{4}{6.4} = \frac{4}{6.4} = \frac{10}{16} = \frac{5}{8} \]
Thus:
\[ \beta_{blue} = (2.4 \times 10^{-4}) \cdot \frac{5}{8} \]

### Step 6: Calculate β_blue.
Calculating the value:
\[ \beta_{blue} = 2.4 \times 10^{-4} \cdot 0.625 = 1.5 \times 10^{-4} \, m \]

### Step 7: Calculate the change in fringe width.
The change in fringe width (Δβ) is given by:
\[ \Delta \beta = \beta_{red} - \beta_{blue} \]
Substituting the values:
\[ \Delta \beta = (2.4 \times 10^{-4}) - (1.5 \times 10^{-4}) = 0.9 \times 10^{-4} \, m \]

### Final Answer:
The change in fringe width when switching from red light to blue light is:
\[ \Delta \beta = 0.9 \times 10^{-4} \, m \]

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