Question

Light whose wavelength is 428 nm passes through two slits in a screen. Part A If the separation between the slits is 12×10−5m , what is the angle to the second-order bright fringe above the central bright fringe? Express your answer to two significant figures and include the appropriate units.

Answer 1

The angle to the second-order bright fringe above the central bright fringe in a double-slit experiment with a slit separation of 12×10⁻⁵m and light wavelength of 428 nm is approximately 4.09 degrees.

Step-by-step explanation:

To find the angle θ to the second-order bright fringe in a double-slit interference pattern, we can use the equation d sin θ = nλ, where d is the separation between the slits, n is the order number, and λ is the wavelength of light.

Given:

• Wavelength, λ = 428 nm = 428 × 10⁻⁹ m.
• Slit separation, d = 12 × 10⁻⁵ m.
• Order number, n = 2 (since we're looking for the second-order bright fringe).

Plugging these values into the equation:

d sin θ = nλ

12 × 10⁻⁵ m sin θ = 2 × 428 × 10⁻⁹ m

Now, we solve for θ:

sin θ = ∼× (∼ / ∼)

sin θ = (856 × 10⁻⁹ m) / (12 × 10⁻⁵ m)

θ = sin⁻¹((856 × 10⁻⁹) / (12 × 10⁻⁵))

θ ≈ sin⁻¹(0.0713) ≈ 4.09° (to two significant figures)

Therefore, the angle to the second-order bright fringe above the central bright fringe is approximately 4.09 degrees.