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Light falls on a double slit with slit separation of 2.02 × 10^−6 m, and the first bright fringe is seen at an angle of 16.5° relative to the

central maximum. Find the wavelength of the light.

1 Answer

5 votes

Answer:

approximately 5.76 × 10^−7 meters, or 576 nanometers

Step-by-step explanation:

The location of bright fringes in a double slit experiment is given by the formula:

d * sin(θ) = m * λ

where:

d is the slit separation,

θ is the angle at which the fringe occurs,

m is the order of the fringe (m = 0 for the central maximum, m = 1 for the first bright fringe, m = 2 for the second bright fringe, and so on), and

λ is the wavelength of the light.

We're looking for the wavelength of the light, and we're given that d = 2.02 × 10^−6 m, θ = 16.5°, and m = 1 (since we're looking at the first bright fringe).

Rearranging the formula to solve for λ gives us:

λ = d * sin(θ) / m

We need to make sure that we're working in radians, as that's what the trigonometric functions in most programming and calculation tools expect. There are π radians in 180 degrees, so to convert from degrees to radians, we multiply by π/180. This gives us θ = 16.5° * π/180 = 0.2873 radians.

Substituting the given values into the formula gives us:

λ = (2.02 × 10^−6 m) * sin(0.2873) / 1

λ ≈ 5.76 * 10^-7 m

So the wavelength of the light is approximately 5.76 × 10^−7 meters, or 576 nanometers (since 1 m = 10^9 nm).

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