Bichromatic light of wavelengths `lambda_1= 5000Å and lambda_2=7000Å` are used in YDSE. Then,

Options

A. 14th order maxima of `lambda_1` will coincide with 10th order maxima of `lambda_2`
B. 14th order maxima of `lambda_1` will coincide with 10th order maxima of `lambda_2`
C. 11th order minima of `lambda_1 will coincide with 8th order minima of `lambda_2`
D. 14th order maxima of `lambda_1` will coincide with 10th order maxima of `lambda_2`

Explanation

To solve the problem of determining which maxima and minima coincide in a Young's Double Slit Experiment (YDSE) using bichromatic light of wavelengths \( \lambda_1 = 5000 \, \text{Å} \) and \( \lambda_2 = 7000 \, \text{Å} \), we will follow these steps:

### Step 1: Understand the Condition for Maxima
The condition for constructive interference (maxima) in YDSE is given by:
\[ n \lambda = d \sin \theta \]
For small angles, this can be approximated as:
\[ n \lambda = \frac{n \lambda D}{d} \]
where \( D \) is the distance from the slits to the screen, and \( d \) is the distance between the slits.

### Step 2: Set Up the Equation for Maxima
For two wavelengths, the condition for maxima can be expressed as:
\[ n_1 \lambda_1 = n_2 \lambda_2 \]
Substituting the values of \( \lambda_1 \) and \( \lambda_2 \):
\[ n_1 \cdot 5000 = n_2 \cdot 7000 \]

### Step 3: Solve for Integer Values of \( n_1 \) and \( n_2 \)
Rearranging gives:
\[ \frac{n_1}{n_2} = \frac{7000}{5000} = \frac{7}{5} \]
This means that for every 7 orders of maxima for \( \lambda_1 \), there are 5 orders of maxima for \( \lambda_2 \).

### Step 4: Find Specific Values
To find specific values of \( n_1 \) and \( n_2 \) that coincide, we can set:
\[ n_1 = 7k \quad \text{and} \quad n_2 = 5k \]
for some integer \( k \).

### Step 5: Example Calculation
Let’s take \( k = 1 \):
- \( n_1 = 7 \)
- \( n_2 = 5 \)

This means the 7th maxima of \( \lambda_1 \) coincides with the 5th maxima of \( \lambda_2 \).

### Step 6: Understand the Condition for Minima
The condition for destructive interference (minima) is given by:
\[ (2n - 1) \frac{\lambda}{2} = d \sin \theta \]
For two wavelengths, the condition for minima can be expressed as:
\[ (2n_1 - 1) \lambda_1 = (2n_2 - 1) \lambda_2 \]

### Step 7: Set Up the Equation for Minima
Substituting the values of \( \lambda_1 \) and \( \lambda_2 \):
\[ (2n_1 - 1) \cdot 5000 = (2n_2 - 1) \cdot 7000 \]

### Step 8: Solve for Integer Values of \( n_1 \) and \( n_2 \)
Rearranging gives:
\[ \frac{(2n_1 - 1)}{(2n_2 - 1)} = \frac{7000}{5000} = \frac{7}{5} \]
This means that for every 7 orders of minima for \( \lambda_1 \), there are 5 orders of minima for \( \lambda_2 \).

### Step 9: Find Specific Values
To find specific values of \( n_1 \) and \( n_2 \) that coincide, we can set:
\[ 2n_1 - 1 = 7k \quad \text{and} \quad 2n_2 - 1 = 5k \]
for some integer \( k \).

### Step 10: Example Calculation
Let’s take \( k = 1 \):
- \( 2n_1 - 1 = 7 \) → \( n_1 = 4 \)
- \( 2n_2 - 1 = 5 \) → \( n_2 = 3 \)

This means the 4th minima of \( \lambda_1 \) coincides with the 3rd minima of \( \lambda_2 \).

### Summary
- The 7th maxima of \( \lambda_1 \) coincides with the 5th maxima of \( \lambda_2 \).
- The 4th minima of \( \lambda_1 \) coincides with the 3rd minima of \( \lambda_2 \).