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Step-by-Step Solution
Step 1: State the Known Parameters
• Slit separation, $d = 0.1\text{ mm}$.
• Bright fringe observed at angle $\theta = \frac{1}{40}$ (in radians, approximately).
• Two wavelengths, $\lambda_1$ and $\lambda_2$, produce a bright fringe at the same angle.
Step 2: Write the Path Difference Condition
For a bright fringe in Young’s double slit experiment, the path difference must be an integer multiple of the wavelength. Mathematically:
$ \Delta = d \sin \theta \approx d \,\theta $.
Since $\theta$ is small, $\sin \theta \approx \theta$.
Step 3: Calculate the Path Difference
$ \Delta = d \,\theta = 0.1\text{ mm} \times \frac{1}{40} $.
Note that:
$ 0.1\text{ mm} = 0.1 \times 10^{-3}\text{ m} = 10^{-4}\text{ m}, $
so
$ \Delta = 10^{-4} \times \frac{1}{40}\text{ m} = \frac{10^{-4}}{40}\text{ m} = 2.5 \times 10^{-6}\text{ m} $.
Converting this to nanometers ($1\text{ m} = 10^9\text{ nm}$):
$ \Delta = 2.5 \times 10^{-6}\text{ m} \times 10^9\text{ nm/m} = 2500\text{ nm}. $
Step 4: Apply the Bright Fringe Condition for Two Wavelengths
Let $n$ and $m$ be the integer orders of the bright fringe for the two wavelengths $\lambda_1$ and $\lambda_2$, respectively. Then,
$ \Delta = n \lambda_1 = m \lambda_2. $
Since $\Delta = 2500\text{ nm}$, we have:
$ n \lambda_1 = 2500\text{ nm}, \quad m \lambda_2 = 2500\text{ nm}. $
Step 5: Find Suitable Integer Multiples Within the Visible Range
We look for integer pairs $(n, \lambda_1)$ and $(m, \lambda_2)$ such that both $\lambda_1$ and $\lambda_2$ lie within the visible range (380 nm to 740 nm). A possible solution is:
If $n = 4$, then $\lambda_1 = \frac{2500}{4} = 625\text{ nm}$.
If $m = 5$, then $\lambda_2 = \frac{2500}{5} = 500\text{ nm}$.
Both 625 nm and 500 nm fall within the visible range.
Step 6: State the Final Answer
Therefore, the wavelengths that satisfy the given conditions are:
$ \lambda_1 = 625\text{ nm} \quad \text{and} \quad \lambda_2 = 500\text{ nm}. $
