To solve the problem of finding the total number of dark fringes due to two wavelengths (420 nm and 540 nm) in a double slit experiment, we can follow these steps:
### Step 1: Understand the condition for dark fringes
Dark fringes in a double slit experiment occur when the path difference between the two waves is an odd multiple of half the wavelength. Mathematically, this can be expressed as:
\[ \Delta x = (2n - 1) \frac{\lambda}{2} \]
where \( n \) is an integer (0, 1, 2, ...), and \( \lambda \) is the wavelength of the light.
### Step 2: Convert given values to consistent units
- Wavelengths:
- \( \lambda_1 = 420 \, \text{nm} = 420 \times 10^{-9} \, \text{m} \)
- \( \lambda_2 = 540 \, \text{nm} = 540 \times 10^{-9} \, \text{m} \)
- Slit separation:
- \( d = 0.0756 \, \text{mm} = 0.0756 \times 10^{-3} \, \text{m} \)
- Distance to screen:
- \( D = 1 \, \text{m} \)
### Step 3: Calculate the maximum path difference for dark fringes
The maximum path difference that can occur on the screen is equal to the slit separation \( d \). Thus, we need to find the maximum value of \( n \) for both wavelengths such that:
\[ (2n - 1) \frac{\lambda}{2} < d \]
### Step 4: Calculate for each wavelength
**For \( \lambda_1 = 420 \, \text{nm} \):**
\[
(2n - 1) \frac{420 \times 10^{-9}}{2} < 0.0756 \times 10^{-3}
\]
\[
(2n - 1) < \frac{0.0756 \times 10^{-3}}{210 \times 10^{-9}} = \frac{0.0756 \times 10^{-3}}{2.1 \times 10^{-7}} \approx 359.05
\]
Thus, the maximum integer value of \( 2n - 1 \) is 359, which gives:
\[
n \leq 180
\]
**For \( \lambda_2 = 540 \, \text{nm} \):**
\[
(2m - 1) \frac{540 \times 10^{-9}}{2} < 0.0756 \times 10^{-3}
\]
\[
(2m - 1) < \frac{0.0756 \times 10^{-3}}{270 \times 10^{-9}} = \frac{0.0756 \times 10^{-3}}{2.7 \times 10^{-7}} \approx 279.63
\]
Thus, the maximum integer value of \( 2m - 1 \) is 279, which gives:
\[
m \leq 140
\]
### Step 5: Calculate the total number of dark fringes
The total number of dark fringes for each wavelength can be calculated as:
- For \( \lambda_1 \): \( n = 180 \) gives 180 dark fringes.
- For \( \lambda_2 \): \( m = 140 \) gives 140 dark fringes.
### Step 6: Combine the results
The total number of dark fringes on the screen due to both wavelengths is:
\[
\text{Total dark fringes} = n + m = 180 + 140 = 320
\]
### Final Answer
The total number of dark fringes due to both wavelengths on the screen is **320**.
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