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

y = 3.69 10⁻² m = 3.69 cm

Step-by-step 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