To find the wavelength of the light, we can use the formula for the position of the interference fringes in a double-slit experiment:
\[ y = \frac{m \lambda L}{d} \]
where:
- \( y \) is the distance from the central maximum to the m-th order maximum (in meters).
- \( m \) is the order of the fringe (for first order, \( m = 1 \)).
- \( \lambda \) is the wavelength of the light (in meters).
- \( L \) is the distance from the slits to the screen (in meters).
- \( d \) is the distance between the slits (in meters).
Given:
- \( y = 16 , \text{mm} = 0.016 , \text{m} \)
- \( m = 1 \) (first order fringe)
- \( L = 0.700 , \text{m} \)
- \( d = 0.0190 , \text{mm} = 0.0000190 , \text{m} \)
We want to find \( \lambda \). Rearranging the formula for \( \lambda \):
\[ \lambda = \frac{y \cdot d}{m \cdot L} \]
Substituting the values into the equation:
\[ \lambda = \frac{0.016 , \text{m} \cdot 0.0000190 , \text{m}}{1 \cdot 0.700 , \text{m}} \]
Calculating the numerator:
\[ 0.016 , \text{m} \cdot 0.0000190 , \text{m} = 0.000000304 , \text{m}^2 \]
Then dividing by \( L \):
\[ \lambda = \frac{0.000000304 , \text{m}^2}{0.700 , \text{m}} = 0.0000004342857 , \text{m} \]
Thus,
\[ \lambda \approx 4.34 \times 10^{-7} , \text{m} \]
The closest choice available is:
4.34×10−7 m
1 answer