Question

In a Young experiment two slits are speared by 6(um), the third dark fringe is formed at an angle 5.6°. The distance between slits and viewing screen is 2 (m).

A- What is the frequency of light used for this experiment?
B- What is the distance between second bright fringe and central fringe?

Answer 1

A) f = 1.28x10¹⁵ Hz

B) y = 0.20 m

Step-by-step explanation:

A) The frequency of light can be found as follows:

`f = (c)/(\lambda)`

Where:

c: is the speed of light = 3.0x10⁸ m/s

λ: is the wavelength

The wavelength can be calculated using the following equation:

`sin(\theta) = (\lambda(m - 1/2))/(d)`

Where:

m = 3, d = 6 μm, θ = 5.6°

`\lambda = (d*sin(\theta))/(m - 1/2) = (6 \cdot 10^(-6) m*sin(5.6))/(3 - 1/2) = 2.34 \cdot 10^(-7) m`

Now, the frequency is:

`f = (c)/(\lambda) = (3.0 \cdot 10^(8) m/s)/(2.34 \cdot 10^(-7) m) = 1.28 \cdot 10^(15) Hz`

Hence, the frequency of light used for this experiment is 1.28x10¹⁵ Hz.

B) The distance between the second bright fringe and central fringe (y) is:

`tan(\theta) = (y)/(D)`

`y = D*tan(\theta) = 2 m*tan(5.6) = 0.20 m`

Therefore, the distance between the second bright fringe and central fringe is 0.20 m.

I hope it helps you!