In Young's double-slit experiment, the slit separation is 0.5 mm and the screen is 0.5 m away from the slit. For a monochromatic light of wavelength 500 nm, the distance of 3rd maxima from the 2nd minima on the other side of central maxima is

Options

A. 2.75 mm
B. 2.5 mm
C. 22.5 mm
D. 2.25 mm

Explanation

To solve the problem, we need to find the distance of the 3rd maxima from the 2nd minima on the other side of the central maxima in Young's double-slit experiment. Here are the steps to arrive at the solution:

### Step 1: Understand the setup
In Young's double-slit experiment, we have:
- Slit separation (d) = 0.5 mm = 0.5 × 10^(-3) m
- Distance from the slits to the screen (D) = 0.5 m
- Wavelength of light (λ) = 500 nm = 500 × 10^(-9) m

### Step 2: Calculate the fringe width (β)
The fringe width (β) is given by the formula:
\[ \beta = \frac{\lambda D}{d} \]
Substituting the values:
\[ \beta = \frac{(500 \times 10^{-9} \text{ m})(0.5 \text{ m})}{0.5 \times 10^{-3} \text{ m}} \]

### Step 3: Simplify the calculation
Calculating the above expression:
\[ \beta = \frac{(500 \times 10^{-9})(0.5)}{0.5 \times 10^{-3}} = \frac{250 \times 10^{-9}}{0.5 \times 10^{-3}} = \frac{250 \times 10^{-9}}{0.5 \times 10^{-3}} = 0.5 \times 10^{-3} \text{ m} = 0.5 \text{ mm} \]

### Step 4: Identify the positions of maxima and minima
- The position of the n-th maxima from the central maximum is given by:
\[ y_n = n \beta \]
- The position of the m-th minima is given by:
\[ y_m = \left(m + \frac{1}{2}\right) \beta \]

### Step 5: Calculate the position of the 3rd maxima and the 2nd minima
- For the 3rd maxima (n = 3):
\[ y_3 = 3 \beta = 3 \times 0.5 \text{ mm} = 1.5 \text{ mm} \]
- For the 2nd minima (m = 2):
\[ y_2 = \left(2 + \frac{1}{2}\right) \beta = \left(2.5\right) \beta = 2.5 \times 0.5 \text{ mm} = 1.25 \text{ mm} \]

### Step 6: Calculate the distance from the 2nd minima to the 3rd maxima
The distance (Δx) from the 2nd minima to the 3rd maxima is given by:
\[ \Delta x = y_3 + y_2 = 1.5 \text{ mm} + 1.25 \text{ mm} = 2.75 \text{ mm} \]

### Final Answer
The distance of the 3rd maxima from the 2nd minima on the other side of the central maxima is **2.75 mm**.

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