Challenge — optical double slit

Problem

Two slits $d = 0{,}2\;\mathrm{mm}$ apart produce interference fringes on a screen at $L = 2\;\mathrm{m}$. With light of $\lambda = 600\;\mathrm{nm}$, calculate the spacing between consecutive fringes.

Solution

Setup. In Young’s experiment, the spacing between consecutive fringes (fringe spacing) on a distant screen is

$\Delta y = \frac{\lambda L}{d}$

Data in SI units. $d = 0{,}2\;\mathrm{mm} = 2\cdot 10^{-4}\;\mathrm{m}$; $L = 2\;\mathrm{m}$; $\lambda = 600\;\mathrm{nm} = 6\cdot 10^{-7}\;\mathrm{m}$.

Substitution.

$\Delta y = \frac{(6\cdot 10^{-7})\cdot 2}{2\cdot 10^{-4}} = \frac{1{,}2\cdot 10^{-6}}{2\cdot 10^{-4}} = 6\cdot 10^{-3}\;\mathrm{m}$

$\ev{\Delta y = 6\;\mathrm{mm}}$

The fringes, though arising from slits fractions of a millimetre apart and light of submicrometric wavelength, end up $6\;\mathrm{mm}$ apart on the screen: it is the large lever arm $L/d$ that makes the interference visible to the naked eye.

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Topics: Ottica Concepts: Interferenza Skills: Principio di sovrapposizione