Answer to Question #304788 in Physics for Koke

Question

Light of wavelenght 500nm is directed normally to a double slit of 1.5mm. If the fringes is observed at a distance of 2m, what is the fringe separation? How many fringes can be observed in 1cm?

Answer 1

The separation is given as follows (see https://www.khanacademy.org/science/physics/light-waves/interference-of-light-waves/v/youngs-double-slit-problem-solving):

$$y = \dfrac{\lambda D}{d}$$

where $\lambda = 500nm = 500\times 10^{-9}m$, $D = 2m$ and $d = 1.5mm = 1.5\times 10^{-3}m$. Thus, find:

$$y = \dfrac{500\times 10^{-9}m\cdot 2m}{1.5\times 10^{-3}m} \approx 6.7\times 10^{-4}m = 0.67mm$$

Thus, the number of fringes in 1 cm is:

$$N = \dfrac{1cm}{0.67mm} \approx 15$$

Answer. separation: 0.67mm, number of fringes: 15.