Question

Describe Young’s double-slit experiment. Explain the setup and the resulting pattern. You may find it helpful to draw a picture. what does the distance between the resulting dots represent? How does it relate to the distance between the slits of the diffraction grating?

Answer 1

Young's double-slit experiment is an experiment first done by English physicist Young to demonstrate wave character of light by diffraction and interference of light originating from a source and then passing through two slits or holes.

Setup and Pattern
In this experiment, light is sent through a pair of vertical slits which is diffracted on the screen spread out horizontally into a pattern of numerous vertical lines or dots(as these lines are very short). Without diffraction and interference, the light would only make two lines on the screen.

Rays Min Screen (a) (b) (c) Wavefront

The interference pattern for a double slit has an intensity that falls off with angle. The below photograph shows multiple bright and dark lines, or fringes, formed by light passing through a double slit.

--2T

The distance((y₂ - y₁)) between the resulting dots or fringes represents the fringe width.

The waves follow different paths from the slits to a common point on a screen. If the paths taken by the two waves differ ( path difference $\Delta l$ ) by any half-integral number of wavelengths [$\left(1/2\right)\lambda$, $\left(3/2\right)\lambda$, $\left(5/2\right)\lambda$, etc.], then destructive interference occurs. The waves start in phase but arrive out of phase. f the paths taken by the two waves differ by an integral number of wavelengths ($\lambda$, $2\lambda$, $3\lambda$, etc.), then constructive interference occurs. The waves start with out of phase and arrive in phase.

s, Je Al S. Al = d sin e Screen

Path difference by simple trigonometry comes out to be ,

NE dsinf

If
NE dsinf
= integral multiples of $\lambda$ for constructive interference producing bright fringe

d sin 0 = mA, for m = 0, 1, – 1, 2, – 2, ... (constructive).

NE dsinf
= half-integral multiples of $\lambda$ for destructive interference producing dark fringe

d sin 0 = (m +) A, for m = 0, 1, – 1, 2, – 2, ... (destructive),

The distance((y₂ - y₁)) between the resulting dots or fringes represents the fringe width.
y₂ - y₁​​​​​​​ = $\Delta y$

$Sin \theta =\theta= \lambda / d =tan\theta =\Delta y/x$​​​​​​​( from the traingle )

$\therefore \Delta y = x \lambda /d$
which relates the slit width d with fringe width $\Delta y$