If the first minima in Young's double-slit experiment occurs directly in front of one of the slits (distance between slit and screen `D = 12 cm` and distance between slits `d = 5 cm)`, then the wavelength of the radiation used can be

Options

A. 2 cm only
B. 4 cm only
C. `2 cm, (2)/(3)cm, (2)/(5) cm`
D. `4cm, (4)/(3) cm, (4)/(5) cm `

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Setup
In Young's double-slit experiment, we have two slits (S1 and S2) separated by a distance \( d \) and a screen at a distance \( D \) from the slits. The first minima occurs directly in front of one of the slits.

### Step 2: Identify Given Values
- Distance between the slits \( d = 5 \, \text{cm} \)
- Distance from the slits to the screen \( D = 12 \, \text{cm} \)

### Step 3: Use the Path Difference Formula
The path difference \( \Delta x \) for the first minima is given by:
\[ \Delta x = S_2P - S_1P = \frac{\lambda}{2} \]
where \( S_2P \) and \( S_1P \) are the distances from the slits to the point on the screen.

### Step 4: Apply Pythagorean Theorem
For small angles, we can approximate the distances using the Pythagorean theorem. The distance \( S_2P \) can be expressed as:
\[ S_2P = \sqrt{D^2 + \left(\frac{d}{2}\right)^2} \]
And since the first minima occurs directly in front of one of the slits, we can express \( S_1P \) as:
\[ S_1P = D \]

### Step 5: Calculate the Path Difference
The path difference can be calculated as:
\[ \Delta x = S_2P - S_1P = \sqrt{D^2 + \left(\frac{d}{2}\right)^2} - D \]
Substituting the values:
\[ \Delta x = \sqrt{12^2 + \left(\frac{5}{2}\right)^2} - 12 \]
Calculating \( \left(\frac{5}{2}\right)^2 = 6.25 \):
\[ \Delta x = \sqrt{144 + 6.25} - 12 = \sqrt{150.25} - 12 \]

### Step 6: Calculate the Square Root
Calculating \( \sqrt{150.25} \):
\[ \sqrt{150.25} \approx 12.25 \]
Thus,
\[ \Delta x \approx 12.25 - 12 = 0.25 \, \text{cm} \]

### Step 7: Relate Path Difference to Wavelength
Since \( \Delta x = \frac{\lambda}{2} \):
\[ 0.25 = \frac{\lambda}{2} \]
Thus, solving for \( \lambda \):
\[ \lambda = 0.5 \, \text{cm} = 5 \, \text{mm} \]

### Step 8: Conclusion
The wavelength of the radiation used can be \( 0.5 \, \text{cm} \) or \( 5 \, \text{mm} \).

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