In young's double slit experiment the `n^(th)` red bright band coincides with `(n+1)^(th)` blue bright band. If the wavelength of red and blue lights are `7500 A^(@)` and `5000 A^(@)`, the value of 'n' is

Options

A. 1
B. 2
C. 5
D. 4

Answer

To solve the problem, we need to determine the value of 'n' in Young's double slit experiment where the nth red bright band coincides with the (n+1)th blue bright band. The wavelengths of the red and blue lights are given as 7500 Å and 5000 Å respectively.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have two different wavelengths of light: red (λ_red = 7500 Å) and blue (λ_blue = 5000 Å). The nth red bright band coincides with the (n+1)th blue bright band.

2. **Fringe Width Formula**:
In Young's double slit experiment, the position of bright fringes is given by the formula:
\[ y_n = n \frac{\lambda D}{d} \]
where \(y_n\) is the position of the nth bright fringe, \(λ\) is the wavelength of light, \(D\) is the distance from the slits to the screen, and \(d\) is the distance between the slits.

3. **Setting Up the Equation**:
For red light, the position of the nth bright band is:
\[ y_n^{red} = n \frac{\lambda_{red} D}{d} = n \frac{7500 D}{d} \]
For blue light, the position of the (n+1)th bright band is:
\[ y_{n+1}^{blue} = (n+1) \frac{\lambda_{blue} D}{d} = (n+1) \frac{5000 D}{d} \]

4. **Equating the Positions**:
Since the nth red band coincides with the (n+1)th blue band, we can set the two equations equal to each other:
\[ n \frac{7500 D}{d} = (n+1) \frac{5000 D}{d} \]

5. **Cancelling Common Terms**:
The terms \(D\) and \(d\) can be cancelled from both sides:
\[ n \cdot 7500 = (n + 1) \cdot 5000 \]

6. **Expanding the Equation**:
Expanding the right-hand side gives:
\[ 7500n = 5000n + 5000 \]

7. **Rearranging the Equation**:
Rearranging the equation to isolate \(n\):
\[ 7500n - 5000n = 5000 \]
\[ 2500n = 5000 \]

8. **Solving for n**:
Dividing both sides by 2500:
\[ n = \frac{5000}{2500} = 2 \]

### Final Answer:
The value of \(n\) is \(2\).