Question

Coherent light of frequency 6.34×1014 Hz passes through two thin slits and falls on a screen 82.0 cm away. You observe that the third bright fringe occurs at ± 3.16 cm on either side of the central bright fringeAt what distance from the central bright fringe will the third dark fringe occur?

Answer 1

Answer:

 y = 3.69 10⁻² m  = 3.69 cm

Explanation:

The interference phenomenon is described by the equations.

      d sin θ  = m λ                               for constrictive interference

      d sin θ = (m + ½) λ                       for destructive interference

Where d is the distance between the slits, m is an integer representing the order of interference λ the wavelength

Let's use the relationship

      c = λ f  

      λ = c / f

      λ = 3 10⁸ / 6.34 10¹⁴

      λ = 4.732 10⁻⁷ m

Let's use trigonometry to find the breast

    tan θ  = y / x

In interference experiments the angle is very small

     tan θ  = sin θ  / cos θ  ≈ sin θ  

We replace

    d y / x = m λ

    d y / x = (m + ½)  λ  

Let's reduce the magnitude to the SI system

       x = 82.0 cm (1m / 100 cm) = 0.820 m

       y = 3.16 cm = 0.0316 cm

Let's use the first relationship to find the distance between the slits

      d = m  λ  x / y

      d = 3 4.732 10⁻⁷ 0.820 /0.0316

      d = 368.38 10⁻⁷ m

      d = 3.6838 10⁻⁵ m

Now we can use the relationship for destructive interference and clear the distance and

        d y / x = (m + ½)  λ

        y = (m + ½)  λ  x / d

       y = (3 + ½) 4,732 10⁻⁷ 0.82 / 3.68 10⁻⁵

       y = 3.69 10⁻² m

       y = 3.69 cm