The Young's double slit experiment is performed with blue light and green light of wavelength `4360 Å` and `5460 Å` respectively. If y is the distance of `4th` maxima from the central one, then

Options

A. `x("blue")=x("green")`
B. `x("blue") gt x("green")`
C. `x("blue") lt x("green")`
D. `(x("blue"))/(x("blue"))=5460/4360`

Explanation

To solve the problem, we need to find the distance of the 4th maxima from the central maximum in a Young's double slit experiment for both blue and green light.

### Step-by-Step Solution:

1. **Identify the Wavelengths:**
- Wavelength of blue light, \( \lambda_B = 4360 \, \text{Å} = 4360 \times 10^{-10} \, \text{m} \)
- Wavelength of green light, \( \lambda_G = 5460 \, \text{Å} = 5460 \times 10^{-10} \, \text{m} \)

2. **Understand the Formula for Fringe Width:**
- The fringe width \( \beta \) in a Young's double slit experiment is given by:
\[ \beta = \frac{\lambda D}{d} \]
where \( D \) is the distance from the slits to the screen, and \( d \) is the distance between the slits.

3. **Calculate the Fringe Width for Blue Light:**
- Let \( \beta_B \) be the fringe width for blue light:
\[ \beta_B = \frac{\lambda_B D}{d} = \frac{4360 \times 10^{-10} D}{d} \]

4. **Calculate the Fringe Width for Green Light:**
- Let \( \beta_G \) be the fringe width for green light:
\[ \beta_G = \frac{\lambda_G D}{d} = \frac{5460 \times 10^{-10} D}{d} \]

5. **Determine the Relationship Between Fringe Widths:**
- Since \( \lambda_B < \lambda_G \), it follows that:
\[ \beta_B < \beta_G \]
This means the fringe width for blue light is less than that for green light.

6. **Find the Position of the 4th Maxima:**
- The position of the \( n^{th} \) maxima from the central maximum is given by:
\[ y_n = n \beta \]
- For blue light (4th maxima):
\[ y_{4B} = 4 \beta_B = 4 \left(\frac{4360 \times 10^{-10} D}{d}\right) \]
- For green light (4th maxima):
\[ y_{4G} = 4 \beta_G = 4 \left(\frac{5460 \times 10^{-10} D}{d}\right) \]

7. **Conclusion:**
- The distance of the 4th maxima from the central one for blue light is:
\[ y_{4B} = \frac{4 \times 4360 \times 10^{-10} D}{d} \]
- The distance of the 4th maxima for green light is:
\[ y_{4G} = \frac{4 \times 5460 \times 10^{-10} D}{d} \]