As you may know, one of the weird things about light is that some of its properties can be explained only by treating it as a wave, while others can be explained only by treating it as a particle. The classical physics that we have applied until now deals only with the particle properties of light. We will now take a look at some phenomena that can only be explained with a wave model of light. These are the phenomenons where wave optics come into play.
Young’s Double-Slit Experiment
The wave theory of light came to prominence with Thomas Young’s double-slit experiment, performed in 1801. We mention this because it is often called “Young’s double-slit experiment,” and you’d best know what SAT II Physics means if it refers to this experiment. The double-slit experiment proves that light has wave properties because it relies on the principles of constructive interference and destructive interference, which are unique to waves.
The double-slit experiment involves light being shone on a screen with—you guessed it—two very narrow slits in it, separated by a distance d. A second screen is set up a distance L from the first screen, upon which the light passing through the two slits shines.
Suppose we have coherent light
—that is, light of a single wavelength
, which is all traveling in phase. This light hits the first screen with the two parallel narrow slits, both of which are narrower than
. Since the slits are narrower than the wavelength, the light spreads out and distributes itself across the far screen.
At any point P on the back screen, there is light from two different sources: the two slits. The line joining P to the point exactly between the two slits intersects the perpendicular to the front screen at an angle .
We will assume that the two screens are very far apart—somewhat more precisely, that L
is much bigger than d
. For this reason, this analysis is often referred to as the “far-field approximation.” This approximation allows us to assume that angles
, formed by the lines connecting each of the slits to P
, are both roughly equal to
. The light from the right slit—the bottom slit in our diagram—travels a distance of l
more than the light from the other slit before it reaches the screen at the point P
As a result, the two beams of light arrive at P
out of phase by d
. If d
= (n + 1/2
, where n
is an integer, then the two waves are half a wavelength out of phase and will destructively interfere. In other words, the two waves cancel each other out, so no light hits the screen at P
. These points are called the minima
of the pattern.
On the other hand, if d sin = n, then the two waves are in phase and constructively interfere, so the most light hits the screen at these points. Accordingly, these points are called the maxima of the pattern.
Because the far screen alternates between patches of constructive and destructive interference, the light shining through the two slits will look something like this:
Note that the pattern is brightest in the middle, where = 0. This point is called the central maximum. If you encounter a question regarding double-slit refraction on the test, you’ll most likely be asked to calculate the distance x between the central maximum and the next band of light on the screen. This distance, for reasons too involved to address here, is a function of the light’s wavelength (), the distance between the two slits (d), and the distance between the two screens (L):
Diffraction is the bending of light around obstacles: it causes interference patterns such as the one we saw in Young’s double-slit experiment. A diffraction grating
is a screen with a bunch of parallel slits, each spaced a distance d
apart. The analysis is exactly the same as in the double-slit case: there are still maxima at d
and minima at d
= (n + 1/2
. The only difference is that the pattern doesn’t fade out as quickly on the sides.
You may also find single-slit diffraction on SAT II Physics. The setup is the same as with the double-slit experiment, only with just one slit. This time, we define d as the width of the slit and as the angle between the middle of the slit and a point P.
Actually, there are a lot of different paths that light can take to P—there is a path from any point in the slit. So really, the diffraction pattern is caused by the superposition of an infinite number of waves. However, paths coming from the two edges of the slit, since they are the farthest apart, have the biggest difference in phase, so we only have to consider these points to find the maxima and the minima.
Single-slit diffraction is nowhere near as noticeable as double-slit interference. The maximum at n = 0 is very bright, but all of the other maxima are barely noticeable. For this reason, we didn’t have to worry about the diffraction caused by both slits individually when considering Young’s experiment.
Light is a transverse wave, meaning that it oscillates in a direction perpendicular to the direction in which it is traveling. However, a wave is free to oscillate right and left or up and down or at any angle between the vertical and horizontal.
Some kinds of crystals have a special property of polarizing light, meaning that they force light to oscillate only in the direction in which the crystals are aligned. We find this property in the crystals in Polaroid disks.
The human eye can’t tell the difference between a polarized beam of light and one that has not been polarized. However, if polarized light passes through a second Polaroid disk, the light will be dimmed the more that second disk is out of alignment with the first. For instance, if the first disk is aligned vertically and the second disk is aligned horizontally, no light will pass through. If the second disk is aligned at a 45ºangle to the vertical, half the light will pass through. If the second disk is also aligned vertically, all the light will pass through.
Wave Optics on SAT II Physics
SAT II Physics will most likely test your knowledge of wave optics qualitatively. That makes it doubly important that you understand the physics going on here. It won’t do you a lot of good if you memorize equations involving d sin but don’t understand when and why interference patterns occur.
One of the more common ways of testing wave optics is by testing your familiarity with different terms. We have encountered a number of terms—diffraction, polarization, reflection, refraction, interference, dispersion—all of which deal with different manipulations of light. You may find a question or two that describe a certain phenomenon and ask which term explains it. Wave Optics is not tough if tackled with proper planning and dedication.
SOLVED EXAMPLES – WAVE OPTICS
MULTIPLE CHOICE QUESTIONS:
|| 1. Which of the following phenomena does NOT affect the direction of a wave of light?
- Consider a light beam incident from air to glass at Brewster’s angle as shown in Fig. 10.1. A Polaroid is placed in the path of the emergent ray at point P and rotated about an axis passing through the center and perpendicular to the plane of the polaroid.
- For a particular orientation there shall be darkness as observed through the Polaroid
- The intensity of light as seen through the Polaroid shall be independent of the rotation
- The intensity of light as seen through the Polaroid shall go through a minimum but not zero for two orientations of the Polaroid
- The intensity of light as seen through the Polaroid shall go through a minimum for four orientations of the Polaroid
- In a Young’s slit double slit experiment, the source is white light. One of the holes is covered by red filter and another by a blue filter. In this case
- There shall be alternate interference patterns of red and blue
- There shall be an interference pattern for red distinct from that for blue
- There shall be no interference fringes
- There shall be an interference pattern for red mixing with one for blue
- The figure shows a standard two slit arrangement with slits S1, S2 and S3. P1, P2 are the two minima points on either side of P. At P2 on the screen, and behind P2 is a second 2-slit arrangement with slits S3, S4 and second screen behind them.
- There would be no interference pattern on the second screen but it would be lighted.
- The second screen would be totally dark.
- There would be a single bright point on the side screen.
- There would be a regular two slit pattern on the second screen
- Two sources S1 and S2 of intensity I1 and I2 are placed in the front of a screen (Fig.(i)). The pattern of intensity distribution seen in the central portion is given by(Fig (ii)). In this case, which of the following statements are true.
- S1 and S2 have same intensities
- S1 and S2 have a constant phase difference
- S1 and S2 have the same phase
- S1 and S2 have the same wavelength
- Consider sunlight incident on a pinhole of width 1000A. The image of the pinhole seen on a screen shall be
- A sharp white ring
- Different from a geometrical image
- A diffused central spot, white in colour
- Diffused coloured region around a sharp central white spot
- Consider the diffraction pattern for a small pinhole. As the size of the hole is increased
- The size decreases
- Intensity increases
- The size increases
- The intensity decreases
- For light diverging from a point source
- The wavefront is spherical
- The intensity decreases in proportion to the distance squared
- The wavefront is parabolic
- The intensity at the wavefront does not depend on the distance
VERY SHORT ANSWER TYPE QUESTIONS
- Is Huygens principle valid for longitudinal sound waves?
- Consider a point at the focal point of the convergent lens. Another convergent lens of short focal length is placed on the other side. What is the nature of the wavefronts emerging from the final image?
- What is the shape of the wavefront on earth for sunlight?
- Why is the diffraction of sound waves more evident in daily experience than that of light wave?
- The human eye has an approzimate angular resolution of 0.0058 rad and a typical photo printer prints a minimum of 300 dpi (dots per inch, 1 inch = 2.54cm). At what minimal distance z should a printed page be held so that one does not see the individual dots?
- A Polaroid (I) is placed in front of a monochromatic source. Another Polaroid (II) is placed in front of this Polaroid (I) and rotated till no light passes. A third Polaroid (III) is now placed placed in between (I) and (II). In this case, will light emerge from (II). Explain.
SHORT ANSWER TYPE QUESTIONS
- Can reflection result in plane polarised light if the light is incident on the interface from the side with higher refractive index?
- For the same objective, find the ratio of the least separation between two points to be distinguished by a microscope for a for a light of 5000 Angstrom and electrons accelerated through 100V used as the illuminating substance.
- Consider a two slit interference arrangements (Fig (i)) such that distance of the screen from the slits is half the distance between the slits. Obtain the value of D in terms of h such that the first minima on the screen falls at a distance D from the centre O.
LONG ANSWER TYPE QUESTIONS
- Figure (i) below shows a two slit arrangement with a source which emits unpolarised light. P is a polariser with axis whose direction is not given. If I0 is the intensity of the principal maxima when no polariser is present, calculate in the present case, the intensity of the principal maxima as well as of the first maxima.
- AC=CO=D, S1C = S2C = d <<D
A small transparent slab containing material of u=1.5 is placed along AS2 (Fig (i)). What will be the distance from O of the principal maxima and of the first minima on either side of the principal maxima obtained in the absence of the glass slab.
- Four identical monochromatic sources A,B,C,D as shown in the fig (i) produce waves of the same wavelength h and are coherent. Two receiver R1 and R2 are at great but equal distances from B.
- Which of the two receivers picks up the larger signal?
- Which of the two receivers picks up the larger signal when B is turned off?
- Which of the two receivers picks up the larger signal when D is turned off?
- Which of the two receivers can distinguish which of the sources B or D has been turned off?
- To ensure almost 100 percent transmissivity, phoographic lenses are often coated with a thin layer of dielectric material. The refractive index of this material is intermediated between that of air and glass (which makes the optical element of the lens). A typically used dielectric film is MgF2 (n=1.38). What should be the thickness of the film so that at the center of the of the visible spectrum (5500 Angstrom) there is maximum transmission.
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