In young's experiment, the separation between 5th maxima and 3rd minima is how many times as that of fringe width ?

Options

A. 5 times
B. 3 times
C. 2.5 times
D. 2 times

Answer

To solve the problem of finding the separation between the 5th maxima and the 3rd minima in Young's experiment in terms of the fringe width, we can follow these steps:

### Step 1: Understand the Definitions
In Young's double-slit experiment, the positions of maxima and minima on the screen can be calculated using the following formulas:
- The position of the nth maxima from the center is given by:
\[ y_m = \frac{n \lambda D}{d} \]
where \( n \) is the order of the maxima, \( \lambda \) is the wavelength, \( D \) is the distance from the slits to the screen, and \( d \) is the distance between the slits.

- The position of the nth minima from the center is given by:
\[ y_n = \frac{(2n - 1) \lambda D}{2d} \]
where \( n \) is the order of the minima.

### Step 2: Calculate the Position of the 5th Maxima
For the 5th maxima (\( n = 5 \)):
\[ y_5 = \frac{5 \lambda D}{d} \]

### Step 3: Calculate the Position of the 3rd Minima
For the 3rd minima (\( n = 3 \)):
\[ y_3 = \frac{(2 \cdot 3 - 1) \lambda D}{2d} = \frac{5 \lambda D}{2d} \]

### Step 4: Find the Separation Between the 5th Maxima and the 3rd Minima
The separation \( S \) between the 5th maxima and the 3rd minima is given by:
\[ S = y_5 - y_3 \]
Substituting the values we calculated:
\[ S = \frac{5 \lambda D}{d} - \frac{5 \lambda D}{2d} \]
To simplify this, we can find a common denominator:
\[ S = \frac{10 \lambda D}{2d} - \frac{5 \lambda D}{2d} = \frac{5 \lambda D}{2d} \]

### Step 5: Relate the Separation to Fringe Width
The fringe width \( \beta \) is defined as:
\[ \beta = \frac{\lambda D}{d} \]
Now, we can express the separation \( S \) in terms of the fringe width:
\[ S = \frac{5 \lambda D}{2d} = \frac{5}{2} \cdot \frac{\lambda D}{d} = \frac{5}{2} \beta \]

### Conclusion
Thus, the separation between the 5th maxima and the 3rd minima is:
\[ \frac{5}{2} \text{ times the fringe width} \]

### Final Answer
The separation between the 5th maxima and the 3rd minima is \( 2.5 \) times the fringe width.

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