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A thin beam of light of wavelength 625 nm goes through a thin slit and falls on a screen 3.00 m past the slit. You observe that the first completely dark fringes occur on the screen at distances of ±8.24 mm from the central bright fringe, and that the central bright fringe has an intensity of 2.00 W/m2 at its center. (a) How wide is the slit?

Respuesta :

Answer:

a = 2.275 10⁻⁴ m

Explanation:

This is a diffraction problem that is described by the equation

         a sin θ = m λ

         

The first dark minimum occurs for m = 1

         a = λ  / sin θ

The angle can be found by trigonometry,

       tan θ = y / x

       θ = tan⁻¹ y / x

Let's reduce the magnitudes to the SI system

        y = 8.24 mm = 8.24 10⁻³ m

        λ  = 625 nm = 625 10⁻⁹ m

       θ = tan⁻¹ 8.24 10⁻³ / 3.00

       θ = 0.002747 rad

We calculate

       a = 625 10⁻⁹ / sin 0.002747

       a = 2.275 10⁻⁴ m

When The screen at distances of ±8.24 mm from the central bright fringe has an intensity of 2.00 W/m2 at its center is a = 2.275 10⁻⁴ m

Computation of Distances

This is a diffraction problem that is described by the equation

Then, a sin θ = m λ  

When The first dark minimum occurs for m = 1

After that,  a = λ  / sin θ

Then, The angle can be found by trigonometry,

tan θ is = y / x

Now, θ = tan⁻¹ y / x

Now, Let's reduce the magnitudes to the SI system

y is = 8.24 mm = 8.24 10⁻³ m

λ is = 625 nm = 625 10⁻⁹ m

θ is = tan⁻¹ 8.24 10⁻³ / 3.00

θ is = 0.002747 rad

We compute

Then, a = 625 10⁻⁹ / sin 0.002747

Therefore, a = 2.275 10⁻⁴ m

Find more information about Distances here:

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