114k views
4 votes
For 600-nm wavelength light and a slit separation of 0.12 mm, what are the angular positions of the first and third maxima in the double slit interference pattern?

a. 0.025° and 0.075°
b. 0.050° and 0.150°
c. 0.075° and 0.225°
d. 0.100° and 0.300°

User KishoreK
by
7.9k points

1 Answer

6 votes

Final answer:

The angular positions of the first and third maxima for 600-nm wavelength light with a slit separation of 0.12 mm are approximately 0.286° and 0.859°, respectively, which, when rounded, are about 0.3° and 0.9°. None of the provided answer choices match these values.

Step-by-step explanation:

To determine the angular positions of the first and third maxima in the double slit interference pattern, we use the formula for interference maxima:

mλ = d sin θ

Where m is the order of the maximum, λ is the wavelength of the light, d is the slit separation, and θ is the angle of the maxima from the central maximum. Plugging in the given values:

For the first maximum (m = 1):
1 * 600 x 10⁻⁹ m = 0.12 x 10⁻3 m * sinθ1
θ1 = sin⁻ⁱ(600 x 10⁻⁹ / 0.12 x 10⁻3) = sin⁻ⁱ(0.005) ≈ 0.286°

For the third maximum (m = 3):
3 * 600 x 10⁻⁹ m = 0.12 x 10⁻3 m * sinθ3
θ3 = sin⁻ⁱ(1800 x 10⁻⁹ / 0.12 x 10⁻3) = sin⁻ⁱ(0.015) ≈ 0.859°

The angles need to be converted to degrees to match the answer choices:

θ1=0.286° is approximately 0.3°, and θ3=0.859° is approximately 0.9°. Therefore, none of the options a, b, c, or d accurately provides the angles for the first and third maxima.

User Cerd
by
7.5k points