To find the wavelength of the light used in the double-slit interference experiment, we can use the formula for the position of the interference fringes:
\[ y_m = \frac{m \lambda L}{d} \]
Where:
- \(y_m\) = position of the m-th order fringe from the central maximum
- \(m\) = order of the fringe (for first order, \(m = 1\))
- \(\lambda\) = wavelength of the light
- \(L\) = distance from the slits to the screen
- \(d\) = distance between the slits
Given:
- \(y_1 = 16 , \text{mm} = 0.016 , \text{m}\) (position of the first order fringe)
- \(L = 0.700 , \text{m}\)
- \(d = 0.0190 , \text{mm} = 0.0000190 , \text{m}\)
Now we can insert these values into the formula and solve for \(\lambda\):
\[ 0.016 = \frac{1 \cdot \lambda \cdot 0.700}{0.0000190} \]
Rearranging the equation to solve for \(\lambda\):
\[ \lambda = \frac{0.016 \cdot 0.0000190}{0.700} \]
Calculating this step-by-step:
- Calculate \(\lambda\): \[ \lambda = \frac{0.016 \cdot 0.0000190}{0.700} = \frac{0.000000304}{0.700} \approx 0.0000004343 , \text{m} \]
- Converting this to nanometers (1 nm = \(10^{-9}\) m): \[ \lambda \approx 434.3 , \text{nm} \]
Thus, the wavelength of the light is approximately:
\[ \lambda \approx 434.3 , \text{nm} \]
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