single-slit diffraction pattern is formed when light of 680.0 nm is passed through a narrow slit. The pattern is viewed on a screen placed one meter from the slit. What is the width of the slit (mm) if the width of the central maximum is 2.69 cm?
Use the single-slit diffraction formula for the angular distance between the two minima on either side of the maximum. If you need a refresher course on that, see
http://en.wikipedia.org/wiki/Diffraction
That angle is given (for small angles) by
theta = 2 (wavelength)/d
Set that angle (in radians) equal to 2.69/100 = 0.0269
Then solve for the slit width, d.
To find the width of the slit, we can use the formula for single-slit diffraction:
sin(θ) = λ / (wavelength)
where θ is the angle of the first minimum (halfway between the central maximum and the first minimum), λ is the wavelength of light, and w is the width of the slit.
By trigonometry, we can approximate sin(θ) as θ, since θ is small. Therefore,
θ = λ / (wavelength)
To find θ, we can use the small angle approximation:
θ = x / L
where x is the width of the central maximum and L is the distance from the slit to the screen.
So,
x / L = λ / (wavelength)
Rearranging the equation, we get:
wavelength = λ * L / x
Now, we can plug in the given values:
wavelength = (680.0 nm) * (1 m) / (2.69 cm)
Convert 2.69 cm to meters: 2.69 cm = 0.0269 m
wavelength = (680.0 nm) * (1 m) / (0.0269 m)
Simplifying, we get:
wavelength = 25.278 m
Therefore, the width of the slit is approximately 25.278 meters.