General Physics II Lab
O1 Two-slit Interference & diffraction grating
General Physics II Lab
O1 Two-slit Interference and Diffraction Grating
Purpose
In this experiment, you will study the interference pattern generated by two closely spaced
slits and use the two slits and multiple slits (a diffraction grating) to determine the
wavelength of visible lights.
Equipment and components
Optics bench, light source, diffraction plate, diffraction grating, diffraction scale, three color
filters (red, green and blue), ray table base and slit mask.
Background
Light is a transverse electromagnetic wave with a wavelength,
, which is directly linked to the
color of visible light. An important experiment to demonstrate the wave nature of light is the
two-slit interference experiment.The essential geometry of the two-slit experiment is shown
in Figure 1. A coherent light passes through the central slot of the diffraction scale and falls
normally on the diffraction plate, which is an opaque screen with two closely spaced, narrow
slits, A and B. According to Huygen’s principle, each slit acts as a new source of light and the
wavelets from each slit will interfere and give rise to an outgoing light, whose intensity varies
with the angle
θ
relative to the incident direction. At the central symmetric point (
θ
= 0), where
the zeroth maxima is located, light rays from slits A and B travel the same distance from the
slits to your eye, so that they are in phase and interfere constructively on your retina. At the
first order maxima (to the left of the viewer) light from slit B travels one wavelength farther
than light from slit A, so that the rays are again in phase and constructive interference occurs
at this position as well. At the nth order maxima, the light from slit B travels n wavelengths
farther than the light from slit A, so again, constructive interference occurs.
λ
Figure 1
Geometry of the two-slit interference experiment
In Fig. 1 the line AC is constructed perpendicular to the line PB. Since the slits are very close
together (not shown to the scale in Fig. 1), lines AP and BP are nearly parallel. Therefore, to
a very close approximation, AP = CP. This means that, for constructive interference to occur
at P, it must be true that BC = n
λ
. From the right triangle ACB, it can be shown that
BC = d sin
θ
’, where d = AB is the distance between the two slits on the diffraction plate.
Furthermore, one can readily show that