Two wavelengths of light `lambda_(1)` and `lambda_(2)` and sent through Young's double-slit apparatus simultaneously. If the third-order bright fringe coincides with the fourth-order bright fringe, then

Options

A. `(lambda_(1))/(lambda_(2)) = (4)/(3)`
B. `(lambda_(1))/(lambda_(2)) = (3)/(4)`
C. `(lambda_(1))/(lambda_(2)) = (5)/(4)`
D. `(lambda_(1))/(lambda_(2)) = (4)/(5)`

Explanation

To solve the problem where two wavelengths of light, λ₁ and λ₂, are sent through Young's double-slit apparatus and the third-order bright fringe of one coincides with the fourth-order bright fringe of the other, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Fringe Position**:
In Young's double-slit experiment, the position of the bright fringes is given by the formula:
\[ y_n = \frac{n \lambda D}{d} \]
where \(y_n\) is the position of the nth order fringe, \(n\) is the order of the 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.

2. **Setting Up the Equations for the Given Wavelengths**:
For the first wavelength \(λ_1\) (third-order fringe):
\[ y_1 = \frac{3 λ_1 D}{d} \]
For the second wavelength \(λ_2\) (fourth-order fringe):
\[ y_2 = \frac{4 λ_2 D}{d} \]

3. **Equating the Positions**:
Since the third-order fringe of \(λ_1\) coincides with the fourth-order fringe of \(λ_2\), we set \(y_1 = y_2\):
\[ \frac{3 λ_1 D}{d} = \frac{4 λ_2 D}{d} \]

4. **Simplifying the Equation**:
We can cancel \(D\) and \(d\) from both sides (assuming \(D\) and \(d\) are not zero):
\[ 3 λ_1 = 4 λ_2 \]

5. **Finding the Ratio of Wavelengths**:
Rearranging the equation gives us:
\[ \frac{λ_1}{λ_2} = \frac{4}{3} \]

### Final Result:
Thus, the ratio of the wavelengths is:
\[ \frac{λ_1}{λ_2} = \frac{4}{3} \]