A double-slit interference pattern is created by two narrow slits spaced 0.21 mm apart. The distance between the first and the fifth minimum on a screen 55 cm behind the slits is 6 mm. What is the wavelength of the light used in this experiment?
Having trouble setting this up.
Thanks
To solve this problem, we can use the formula for the double-slit interference pattern given by:
d*sin(θ) = m*λ
where:
- d is the distance between the slits (0.21 mm),
- θ is the angle between the center, the first minimum, and the center, the fifth minimum,
- m is the order of the minimum (in this case, m = 1 since we are considering the first minimum),
- and λ is the wavelength of the light.
Given that the distance between the first and fifth minimum on the screen is 6 mm and the distance between the screen and the slits is 55 cm, we can use the small-angle approximation to find the value of θ.
The small-angle approximation states that for small angles, sin(θ) ≈ θ (in radians).
So, we can rewrite the formula as:
d*θ = m*λ
To find θ, we divide the distance between the first and fifth minimum (6 mm) by the distance between the screen and the slits (55 cm). We also convert the length to meters:
θ = (6 mm) / (55 cm) = 0.006/0.55 ≈ 0.0109 radians
Substituting this value into the formula, we have:
(0.21 mm) * (0.0109 radians) = (1) * λ
Simplifying, we find:
λ = (0.21 mm) * (0.0109 radians) / (1)
λ = 0.002289 mm
However, it is more common to express wavelength in meters. To convert millimeters to meters, we divide by 1000:
λ = (0.002289 mm) / (1000) = 2.289 × 10^(-6) m
Therefore, the wavelength of the light used in this experiment is approximately 2.289 × 10^(-6) meters.