EXAMPLE Young’s double-slit experiment is performed with 589-nm light and a distance of 2.00 m between the slits and the screen. The tenth interference minimum is observed 7.26 mm from the central maximum. Determine the spacing of the slits.

EXAMPLE Two narrow, parallel slits separated by 0.250 mm are illuminated by green light (λ =546.1 nm). The interference pattern is observed on a screen 1.20 m away from the plane of the slits. Calculate the distance (a) from the central maximum to the first bright region on either side of the central maximum and (b) between the first and second dark bands. Solution

CHANGE OF PHASE DUE TO REFLECTION Another simple, yet ingenious, arrangement for producing an interference pattern with a single light source is known as Lloyd’s mirror. Light waves can reach the viewing point P either by the direct path SP or by the path involving reflection from the mirror. The reflected ray can be treated as a ray originating at the source S´ behind the mirror. Source S´, which is the image of S, can be considered a virtual source.

At points far from the source, an interference pattern due to waves from S and S´ is observed, just as for two real coherent sources. the positions of the dark and bright fringes are reversed relative to the patterns of Young’s experiment. because the coherent sources S and S´ differ in phase by 180°, a phase change produced by reflection. To illustrate the point further.

consider P´, the point where the mirror intersects the screen consider P´, the point where the mirror intersects the screen. This point is equidistant from S and S´. If path difference alone were responsible for the phase difference, a bright fringe would be observed at P (because the path difference is zero for this point.

corresponding to the central fringe of the two-slit interference pattern. Instead, we observe a dark fringe at P, from which we conclude that a 180° phase change must be produced by reflection from the mirror. In general, an electromagnetic wave undergoes a phase change of 180° upon reflection from a medium that has an index of refraction higher than the one in which the wave was travelling.

An analogy can be drawn between reflected light waves and the reflections of a transverse wave on a stretched string when the wave meets a boundary, as in Figure when it is reflected from the boundary of a denser string or from a rigid barrier and undergoes no phase change when it is reflected from the boundary of a less dense string. Similarly, an electromagnetic wave undergoes a 180° phase change when reflected from the boundary of a medium with index of refraction higher than the one in which it has been travelling. There is no phase change when the wave is reflected from a boundary leading to a medium of lower index of refraction. The transmitted wave that crosses the boundary also undergoes no phase change.

where λ is the wavelength of light in vacuum. INTERFERENCE IN THIN FILMS Interference effects are commonly observed in thin films, such as the thin surface of a soap bubble or thin layers of oil on water. The varied colours observed when incoherent white light is incident on such films result from the interference of waves reflected from the two surfaces of the film. Assume that the light rays travelling in air are nearly normal to the two surfaces of the film. To determine whether the reflected rays interfere constructively or destructively, we first note the following facts: An electromagnetic wave travelling from a medium of index of refraction n1 toward a medium of index of refraction n2 undergoes a 180° phase change on reflection when n2 > n1. There is no phase change in the reflected wave if n2 < n1. 2. The wavelength of light λn in a medium with index of refraction n is where λ is the wavelength of light in vacuum.

We apply these rules to the film of Figure 24.7 ray 1, which is reflected from the upper surface A, undergoes a phase change of 180° with respect to the incident wave. Ray 2, which is reflected from the lower surface B, undergoes no phase change with respect to the incident wave. Therefore, ray 1 is 180° out of phase with respect to ray 2, which is equivalent to a path difference of λn /2. we must also consider the fact that ray 2 travels an extra distance of 2t before the waves recombine in the air above the surface.

if 2t = λn/2, then rays 1 and 2 recombine in phase, and constructive interference results. In general, the condition for constructive interference in thin films is This condition takes into account two factors: (1) the difference in path length for the two rays (the term mλn) and (2) the 180° phase change upon reflection (the term λn/2). Because λn = λ/n, we can write Equation 24.8 in the form

If the extra distance 2t travelled by ray 2 is a multiple of λn, then the two waves combine out of phase and the result is destructive interference. The general equation for destructive interference in thin films is Equations 24.9 and 24.10 for constructive and destructive interference are valid when there is only one phase reversal.(nair less than nfilm ) If the film is placed between two different media, one of lower refractive index than the film and one of higher refractive index, Equations 24.9 and 24.10 are reversed: