In standard Young's double-slit experiment, 9 fringes are found to be formed in a portion of screen. Wavelength of light used is 600 nm. How many fringes will be formed in the same portion of screen if wavelength of light is changed to 450 nm?

Options

A. 12
B. 18
C. 14
D. 20

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between fringe width and wavelength
In Young's double-slit experiment, the fringe width (β) 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 total width for 9 fringes with the first wavelength
The total width for 9 fringes can be calculated as:
\[ \text{Total width for 9 fringes} = 9 \beta_1 = 9 \left( \frac{\lambda_1 D}{d} \right) \]
Substituting \( \lambda_1 = 600 \, \text{nm} = 600 \times 10^{-9} \, \text{m} \):
\[ \text{Total width for 9 fringes} = 9 \left( \frac{600 \times 10^{-9} D}{d} \right) \]

### Step 3: Set up the equation for the new wavelength
Now, we want to find how many fringes (n) will be formed with the new wavelength \( \lambda_2 = 450 \, \text{nm} = 450 \times 10^{-9} \, \text{m} \):
\[ \text{Total width for n fringes} = n \beta_2 = n \left( \frac{\lambda_2 D}{d} \right) \]
Substituting \( \lambda_2 \):
\[ \text{Total width for n fringes} = n \left( \frac{450 \times 10^{-9} D}{d} \right) \]

### Step 4: Equate the total widths
Since the total width for 9 fringes with the first wavelength is equal to the total width for n fringes with the second wavelength, we can equate the two expressions:
\[ 9 \left( \frac{600 \times 10^{-9} D}{d} \right) = n \left( \frac{450 \times 10^{-9} D}{d} \right) \]

### Step 5: Simplify the equation
Cancel \( D \) and \( d \) from both sides:
\[ 9 \times 600 \times 10^{-9} = n \times 450 \times 10^{-9} \]
This simplifies to:
\[ 9 \times 600 = n \times 450 \]

### Step 6: Solve for n
Now, solve for \( n \):
\[ n = \frac{9 \times 600}{450} \]
Calculating the right side:
\[ n = \frac{5400}{450} = 12 \]

### Conclusion
Thus, the number of fringes formed when the wavelength is changed to 450 nm is **12**.

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