To solve the problem step by step, we will follow the principles of wave optics and the conditions for destructive interference in a double-slit experiment.
### Step 1: Understand the Problem
We have a double slit with a separation \( d = 1.5 \, \text{mm} = 1.5 \times 10^{-3} \, \text{m} \) illuminated by white light. The screen is located at a distance \( D = 120 \, \text{cm} = 1.2 \, \text{m} \). A pinhole is made at a distance \( y = 3.0 \, \text{mm} = 3.0 \times 10^{-3} \, \text{m} \) from the central white fringe. We need to find the second longest wavelength that will be absent in the transmitted light.
### Step 2: Use the Condition for Minima
The condition for destructive interference (minima) in a double-slit experiment is given by:
\[
d \sin \theta = (m + \frac{1}{2}) \lambda
\]
where \( m \) is the order of the minima (0, 1, 2, ...), and \( \lambda \) is the wavelength of light.
### Step 3: Approximate for Small Angles
For small angles, we can use the approximation \( \sin \theta \approx \tan \theta \approx \frac{y}{D} \). Therefore, we can rewrite the equation as:
\[
d \frac{y}{D} = (m + \frac{1}{2}) \lambda
\]
### Step 4: Rearranging for Wavelength
Rearranging the equation to solve for \( \lambda \):
\[
\lambda = \frac{d y}{D (m + \frac{1}{2})}
\]
### Step 5: Substitute the Values
Now we substitute the known values into the equation:
- \( d = 1.5 \times 10^{-3} \, \text{m} \)
- \( y = 3.0 \times 10^{-3} \, \text{m} \)
- \( D = 1.2 \, \text{m} \)
Substituting these values:
\[
\lambda = \frac{(1.5 \times 10^{-3}) (3.0 \times 10^{-3})}{1.2 (m + \frac{1}{2})}
\]
### Step 6: Simplifying the Equation
Calculating the numerator:
\[
1.5 \times 10^{-3} \times 3.0 \times 10^{-3} = 4.5 \times 10^{-6}
\]
Thus,
\[
\lambda = \frac{4.5 \times 10^{-6}}{1.2 (m + \frac{1}{2})}
\]
### Step 7: Calculate for Different Values of \( m \)
Now we calculate \( \lambda \) for different values of \( m \):
- For \( m = 0 \):
\[
\lambda_0 = \frac{4.5 \times 10^{-6}}{1.2 \times 0.5} = \frac{4.5 \times 10^{-6}}{0.6} = 7.5 \times 10^{-6} \, \text{m} = 7500 \, \text{Å}
\]
- For \( m = 1 \):
\[
\lambda_1 = \frac{4.5 \times 10^{-6}}{1.2 \times 1.5} = \frac{4.5 \times 10^{-6}}{1.8} = 2.5 \times 10^{-6} \, \text{m} = 2500 \, \text{Å}
\]
- For \( m = 2 \):
\[
\lambda_2 = \frac{4.5 \times 10^{-6}}{1.2 \times 2.5} = \frac{4.5 \times 10^{-6}}{3.0} = 1.5 \times 10^{-6} \, \text{m} = 1500 \, \text{Å}
\]
- Continue this process until \( m = 10 \).
### Step 8: List the Wavelengths
Continuing to calculate for \( m = 3 \) to \( m = 10 \):
- \( m = 3 \): \( 1071.4 \, \text{Å} \)
- \( m = 4 \): \( 833.3 \, \text{Å} \)
- \( m = 5 \): \( 681.8 \, \text{Å} \)
- \( m = 6 \): \( 571.5 \, \text{Å} \)
- \( m = 7 \): \( 500.0 \, \text{Å} \)
- \( m = 8 \): \( 444.4 \, \text{Å} \)
- \( m = 9 \): \( 394.7 \, \text{Å} \)
### Step 9: Identify Wavelengths in the Range
We need to find the wavelengths that fall within the range of \( 4000 \, \text{Å} \) to \( 8000 \, \text{Å} \):
- \( 7500 \, \text{Å} \) (for \( m = 0 \))
- \( 6818 \, \text{Å} \) (for \( m = 6 \))
- \( 5769 \, \text{Å} \) (for \( m = 7 \))
### Step 10: Conclusion
The second longest wavelength that is absent in the transmitted light is:
\[
\text{Second longest wavelength} = 5769 \, \text{Å}
\]
