In Young's double-slit experiment, two wavelengths of light are used simultaneously where `lambda_(2) = 2 lambda_(1)`. In the fringe pattern observed on the screen,

Options

A. maxima of wavelength `lambda_(2)` can coincide with minima of wavelength `lambda_(1)`
B. fringe width of `lambda_(2)` will be double that of fringe width of `lambda_(1)` and nth order maxima of `lambda_(2)` will coincide with 2nd order maxima of `lambda_(1)`
C. nth order minima of `lambda_(2)` will coincide with 2th order minima of `lambda_(1)`
D. none of above

Explanation

To solve the problem regarding Young's double-slit experiment with two wavelengths of light, where \( \lambda_2 = 2 \lambda_1 \), we will follow these steps:

### Step 1: Understand the fringe width formula
The fringe width (β) in a double-slit experiment is given by the formula:
\[ \beta = \frac{D \lambda}{d} \]
where:
- \( D \) is the distance from the slits to the screen,
- \( \lambda \) is the wavelength of the light used,
- \( d \) is the distance between the slits.

### Step 2: Calculate the fringe widths for both wavelengths
For the two wavelengths, we can express the fringe widths as follows:
1. For \( \lambda_1 \):
\[ \beta_1 = \frac{D \lambda_1}{d} \]
2. For \( \lambda_2 \) (where \( \lambda_2 = 2 \lambda_1 \)):
\[ \beta_2 = \frac{D \lambda_2}{d} = \frac{D (2 \lambda_1)}{d} = 2 \beta_1 \]

### Step 3: Determine the positions of the maxima
The position of the nth order maxima for each wavelength can be expressed as:
1. For \( \lambda_1 \):
\[ y_n = n \frac{D \lambda_1}{d} \]
2. For \( \lambda_2 \):
\[ y_m = m \frac{D \lambda_2}{d} = m \frac{D (2 \lambda_1)}{d} = 2m \frac{D \lambda_1}{d} \]

### Step 4: Find conditions for coinciding maxima
To find the conditions under which the maxima from both wavelengths coincide, we set the positions equal:
\[ n \frac{D \lambda_1}{d} = 2m \frac{D \lambda_1}{d} \]
This simplifies to:
\[ n = 2m \]
This means that for every nth order maximum of \( \lambda_1 \), there is a corresponding mth order maximum of \( \lambda_2 \) when \( m = \frac{n}{2} \).

### Step 5: Conclusion
From the above analysis, we conclude that the maxima from the two wavelengths coincide when \( n \) is even, specifically when \( n = 2m \).