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

Answer:

A) f = 1.28x10¹⁵ Hz

B) y = 0.20 m

Explanation:

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

$f = \frac{c}{\lambda}$

Where:

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

λ: is the wavelength

The wavelength can be calculated using the following equation:

$sin(\theta) = \frac{\lambda(m - 1/2)}{d}$

Where:

m = 3, d = 6 μm, θ = 5.6°        

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

Now, the frequency is:

$f = \frac{c}{\lambda} = \frac{3.0 \cdot 10^{8} m/s}{2.34 \cdot 10^{-7} m} = 1.28 \cdot 10^{15} Hz$    

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:

$tan(\theta) = \frac{y}{D}$        

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

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

I hope it helps you!