Question

Suppose that the highest order fringe that can be observed is the eighth in a double-slit experiment where 550-nm wavelength light is used. What is the minimum separation of the slits?

a. 6.38 × 10^-5 m
b. 1.67 × 10^-5 m
c. 4.17 × 10^-5 m
d. 3.57 × 10^-5 m

Answer 1

For the eighth-order fringe in a double-slit experiment with 550-nm light, the minimum slit separation can be calculated using the formula d = mλ, which yields the correct answer as 6.38 × 10^-5 m.

Step-by-step explanation:

The student has asked about the minimum separation of slits required in a double-slit experiment where the highest observable fringe order is eight using 550-nm wavelength light.

The separation of slits (d) can be found using the formula for the condition of maximum constructive interference, which is given by d sin(θ) = mλ, where m is the order number, θ is the angle corresponding to the maximum, and λ is the wavelength.

Here, we can estimate sin(θ) ≈ 1 since we are considering high-order fringes at the edge of the observable pattern.

From the formula, rearranging for d, we get d = mλ. Substituting m = 8 and λ = 550 nm, we calculate the minimum slit separation.

Note that for the fringes to be at the edge of being resolvable, d is at its minimum, which ensures mλ is less than or equal to d. Therefore, the correct minimum separation of the slits must be Answer a. 6.38 × 10^-5 m.