Question

If u shine a light of frequency 375Hz on a double slit setup, and u measure the slit separation to be 950nm and the screen distance to be 4030 nm away what is the distance from the zero order fringe to the first order fringe

Answer 1

Answer:

Y = 3.39 x 10⁶ m

Explanation:

We will use Young's Double Slit formula here:

$Y = \frac{\lambda L}{d}$

where,

Y = Fringe spacing = ?

λ = wavelength = $\frac{speed\ of\ light}{frequency} = \frac{3\ x\ 10^8\ m/s}{375\ Hz}$ = 8 x 10⁵ m

L = screend distance = 4030 nm = 4.03 x 10⁻⁶ m

d = slit separation = 950 nm = 9.5 x 10⁻⁷ m

Therefore,

$Y = \frac{(8\ x\ 10^5\ m)(4.03\ x\ 10^{-6}\ m)}{9.5\ x\ 10^{-7}\ m}$

Y = 3.39 x 10⁶ m