Answer to Question #166573 in Optics for Ahmed Raza

Question

Explain Youngs double experiment with a monochromatic light source showing the wave model for light. (No more than 200 words)

Answer 1

Introduction:

Young’s double-slit experiment uses two coherent sources of light placed at a small distance apart, usually, only a few orders of magnitude greater than the wavelength of light is used. Young’s double-slit experiment helped in understanding the wave theory of light which is explained with the help of a diagram. A screen or photodetector is placed at a large distance ’D’ away from the slits as shown.

The original Young’s double-slit experiment used diffracted light from a single source passed into two more slits to be used as coherent sources. Lasers are commonly used as coherent sources in the modern-day experiments.

Derivation:

Consider a monochromatic light source ‘s’ kept at a considerable distance from two slits s₁ and s₂. S is equidistant from s₁ and s₂. s₁ and s₂ behave as two coherent sources, as both bring derived from S.

The light passes through these slits and falls on a screen which is at a distance ‘D’ from the position of slits s₁ and s₂. ‘d’ be the separation between two slits.

When the slit separation (d) and the screen distance (D) are kept unchanged, to reach P the light waves from s₁ and s₂ must travel different distances. It implies that there is a path difference in Young’s double slit experiment between the two light waves from s₁ and s₂.

Since D > > d, the two light rays are assumed to be parallel, then the path difference and

 d is a fraction of a millimeter and λ is a fraction of a micrometer for visible light.

Under these conditions θ is small, thus we can use the approximation sin θ $\approx$ tan θ = γ/D

∴ path difference, Δz = γ/D

This is the path difference between two waves meeting at a point on the screen. Due to this path difference in Young’s double slit experiment, some points on the screen are bright and some points are dark.

Position of Fringes In Young’s Double Slit Experiment:

Position of Bright Fringes

For maximum intensity at P

Δz = nλ (n = 0, ±1, ±2, . . . .)

Or d sin θ = nλ (n = 0, ±1, ±2, . . . .)

The bright fringe for n = 0 is known as the central fringe. Higher order fringes are situated symmetrically about the central fringe.

The position of n^th bright fringe is given by $y_{dark}$ = (nλ\d)D (n = 0, ±1, ±2, . . . .)

Position of Dark Fringes

For minimum intensity at P,

$\Delta z=\left( 2n-1 \right)\frac{n\lambda }{2}\left( n=\pm 1,\pm 2,….. \right)$

or $d\sin \theta =\left( 2n-1 \right)\frac{n\lambda }{2}$

The first minima are adjacent to the central maximum on either side. We will obtain the position of dark fringe as

${{y}_{dark}}=\frac{\left( 2n-1 \right)\lambda D}{2d}\left( n=\pm 1,\pm 2,….. \right)$

Fringe Width:

Distance between two adjacent bright (or dark) fringes is called the fringe width.

$\beta z\frac{n\lambda D}{d}-\frac{\left( n-1 \right)\lambda D}{d}=\frac{\lambda D}{d} ⇒ \beta \propto \lambda$

If the apparatus of Young’s double slit experiment is immersed in a liquid of refractive index (u), then wavelength of light and hence fringe width decreases ‘u’ times.

Shape of Interference Fringes in YDSE:

From the given YDSE diagram, the path difference from the two slits is given by

${{s}_{2}}p-{{s}_{1}}p\approx d\sin \theta$

The above equation represents a hyperbola with its two foci as s₁ and s₂.