In Fig. 35-51 a, the waves along rays 1 and 2 are initially...

In Fig. $35-51 a,$ the waves along rays 1 and 2 are initially in phase, with the same wavelength $\lambda$ in air. Ray 2 goes through a material with length $L$ and index of refraction $n .$ The rays are then reflected by mirrors to a common point $P$ on a screen. Suppose that we can vary L from 0 to 2400 nm. Suppose also that, from $L=0$ to $L_{s}=900 \mathrm{nm}$ the intensity $I$ of the light at point $P$ varies with $L$ as given in Fig. $35-52$ . At what values of $L$ greater than $L_{s}$ is intensity $I$ (a) maximum and (b) zero? (c) What multiple of $\lambda$ gives the phase difference between ray 1 and ray 2 at common point $P$ when $L=1200 \mathrm{nm}$ ?

Key Concepts

Fringe (Interference) Conditions

For two interfering waves, the conditions for constructive and destructive interference can be derived from their phase difference. Constructive interference (resulting in intensity maximum) occurs when the phase difference is an integer multiple of 2?. This leads to the condition (n – 1)L = m? for m = 0, 1, 2, … Conversely, destructive interference (zero intensity) occurs when the phase difference is an odd multiple of ?, i.e. (n – 1)L = (m + ½)?. In the context of the problem, knowing one value (for example, that a particular L = Ls corresponds to a zero in intensity) allows one to relate the wavelength ? to the quantity (n – 1) and then predict the values of L for subsequent maxima and minima. For example, if Ls = 900 nm is such that (n – 1)900 = ½?, then (n – 1)? = 1800 nm, so the first maximum for L > Ls occurs when L = 1800 nm and the phase difference at L = 1200 nm corresponds to an optical path difference of (n – 1)1200 = (1200/1800)? = (2/3)?.

Phase Difference

The phase difference between two waves is determined by the difference in their optical path lengths. Specifically, if the optical path difference is ?, then the phase difference is given by ?? = (2?/?)?, where ? is the wavelength in air. In this case, since one ray travels extra distance in the material, the phase difference is ?? = (2?/?)(n – 1)L. This phase difference is what causes the interference pattern, with specific conditions leading to maxima (constructive interference) and minima (destructive interference).

Interference

Interference is the phenomenon that occurs when two or more waves superpose to form a resultant wave of greater, lower, or the same amplitude. In optics, when two light rays are combined, their phase relationship (whether they are in phase or out of phase) determines if they interfere constructively (resulting in maximum intensity) or destructively (resulting in zero intensity). This concept is essential for understanding fringe patterns and the distribution of light intensity in experiments such as the one described.

Optical Path Length

The optical path length (OPL) is the product of the physical distance a light ray travels in a medium and the index of refraction of that medium. It represents the equivalent distance that light would travel in vacuum to acquire the same phase change. In the given problem, one ray travels through a material of length L and index n, so its optical path length is nL. The optical path difference between the two rays (one in air and one in the material) is a key parameter in determining the phase difference between them.

Transcript

Please give Ace some feedback