As an admirer of Thomas Young, you perform a double-slit experiment in his honor. You set your slits $1.03 \mathrm{~mm}$ apart and position your screen $3.93 \mathrm{~m}$ from the slits. Although Young had to struggle to achieve a monochromatic light beam of sufficient intensity, you simply turn on a laser with a wavelength of $631 \mathrm{~nm}$.
How far on the screen are the first bright fringe and the second dark fringe from the central bright fringe? Express your answers in millimeters.
first bright fringe:
$\mathrm{mm}$
second dark fringe:
$\mathrm{mm}$
Answer
The first bright fringe is approximately \(\boxed{2.41 \, \text{mm}}\) from the central bright fringe, and the second dark fringe is approximately \(\boxed{3.61 \, \text{mm}}\) from the central bright fringe.
Step 1 ：We are given the following values: the distance from the slits to the screen, \(L = 3.93 \, \text{m}\), the wavelength of the light, \(\lambda = 631 \, \text{nm} = 631 \times 10^{-9} \, \text{m}\), and the distance between the slits, \(d = 1.03 \, \text{mm} = 1.03 \times 10^{-3} \, \text{m}\).
Step 2 ：We substitute these values into the formula for the position of the first bright fringe, \(y_{bright} = L \times \lambda \times m_{bright} / d\), where \(m_{bright} = 1\).
Step 3 ：We then convert the result from meters to millimeters by multiplying by 1000, giving us the position of the first bright fringe in millimeters.
Step 4 ：We substitute the given values into the formula for the position of the second dark fringe, \(y_{dark} = L \times \lambda \times (m_{dark} + 0.5) / d\), where \(m_{dark} = 1\).
Step 5 ：We then convert the result from meters to millimeters by multiplying by 1000, giving us the position of the second dark fringe in millimeters.
Step 6 ：The first bright fringe is approximately \(\boxed{2.41 \, \text{mm}}\) from the central bright fringe, and the second dark fringe is approximately \(\boxed{3.61 \, \text{mm}}\) from the central bright fringe.
