ASP Physics - Wave Optics
- Path difference: \partial = d \sin\theta
- Constructive interference in Double slit: \partial = d \sin\theta_{bright} = m\lambda, where m = 0, \pm 1, \pm 2, …
- Destructive interference in Double slit: \partial = d \sin\theta_{dark} = (m + \frac{1}{2})\lambda, where m = 0, \pm 1, \pm 2, …
- Fringe location in Double slit experiment: y = L \sin\theta
- Positions of Bright and dark fringes from the center in Double slit:
- y_{bright} = m \frac{\lambda L}{d}
- y_{dark} = (m + \frac{1}{2}) \frac{\lambda L}{d}
- Single – slit diffraction: \sin\theta_{dark} = m \frac{\lambda}{a}, where m = \pm 1, \pm 2, \pm 3 …
- Diffraction Grating: d = \frac{1}{n} cm, where n is the number of lines per cm
- d \sin\theta_{bright} = m\lambda, where m = 0, \pm 1, \pm 2, …
- Polarization: I = I_0 cos^2\theta
Topic 10: Wave Optics
Subtopic 10.1: Interference in Young’s Double Slit Experiment
- KPI: 10.1.1 and 10.1.2
- Interference: Two or more waves can combine to form a new wave. The result of the superposition of two or more waves is called interference.
- Wave interference can be constructive or destructive.
- Waves that are in phase add to give a larger amplitude. This is constructive interference.
- Waves that are half a wavelength out of phase interfere destructively. If the individual waves have equal amplitudes, their sum will have zero amplitude.
- Coherent and Monochromatic Source
- Two waves are coherent if the phase difference between them is constant. For this to be the case they must have the same frequency.
- Monochromatic – means having only one wavelength (or frequency or color) of light present.
- A coherent source is always monochromatic while a monochromatic source may or may not be a coherent source.
- Two separate sources can be practically used as monochromatic sources, but for coherence, two virtual sources designed from a single monochromatic source must be used.
- Conditions for Interference
- The sources of the waves must be coherent, which means they emit identical waves with a constant phase difference.
- The waves should be monochromatic - they should be of a single wavelength (frequency or color)
- Example 1: What are the conditions required for interference of light to occur?
- To set up a stable and clear interference pattern, two conditions must be met:
- The sources of the waves must be coherent, which means they emit identical waves with a constant phase.
- The waves should be monochromatic - they should be of a single wavelength.
- Example 2: Your automobile has two headlights. What sort of interference pattern do you expect to see from them? Why?
- You will not see an interference pattern from the automobile headlights, for two reasons:
- The first is that the headlights are not coherent sources and are therefore incapable of producing sustained interference.
- The headlights are so far apart in comparison to the wavelengths emitted that, even if they were made into coherent sources, the interference maxima and minima would be too closely spaced to be observable.
- Example 2: What are two characteristics of laser light that make it useful?
- Lasers are coherent, monochromatic and directional (narrow beam in a specific direction) and are perfect to produce a clear interference pattern.
- KPI 10.1.3 Young’s Double Slit Experiment
- Thomas Young first demonstrated interference from light waves with a double slit. His experiment showed that monochromatic light that passes through two slits produces an interference pattern of bright and dark fringes.
- The separation between the fringes is directly related to the wavelength of the light.
- The light from a single slit source goes through two very narrow openings at S1 and S2 so that the light waves have identical frequency and phase. The light beam is also considered to be of one color. Each of the slits act as a source for circular expanding waves.
- The points of intersection of two crests, one from each slit, are points of constructive interference (bright).
- The point of intersection of a crest from one slit and a trough from the other slit is a point of destructive interference (dark).
- The interference pattern called fringes, consisting of alternating bright and dark bars, will be seen on the screen.
- Constructive interference occurs at point P when the waves combine. Because both waves travel the same distance, they arrive at P in phase, forming a bright fringe.
- Constructive interference also occurs at point Q. Although the two waves start in phase, the upper wave has to travel one wavelength farther than the lower wave to reach Q and so they still arrive in phase. So a second bright fringe is appears at this location.
- Destructive interference occurs at R when the two waves combine. Between points P and Q the upper wave falls half a wavelength behind the lower wave. So a dark fringe is appears at this location.
- Example 1: Which will be produced when blue light with a wavelength of 4.7 × 10^{-7} meters passes through a double slit?
- C. alternate blue and black hands
- Example 2: Two beams of coherent light are shining on the same piece of white paper. With respect to the crests and troughs of such waves, darkness will occur on the paper where:
- B. the crest from one wave overlaps with the trough from the other
- KPI 10.1.4 Consider the geometric construction below for describing Young’s double-slit experiment.
- The light falls on the screen at the point P which is at a distance y from the center O.
- The distance between the double-slit system and the screen is L
- The two slits are separated by the distance d
- Distance travelled by the light ray from slit S1 to point P on the screen is r1
- Distance travelled by the light ray from slit S2 to point P on the screen is r2
- Thus, the light ray from slit S2 travels an extra distance of \partial = r2 - r1 than light ray from slit S1.
- This extra distance is termed as Path difference (∂).
- When we assume that r1 is parallel to r2, the path difference (∂) between the two rays is: \partial = d \sin \theta = r2 - r1
- For this approximation to be valid, it is essential that L >> d
- KPI 10.1.5 and 10.1.6 Conditions for Bright Fringes (Constructive Interference) in the two-slit experiment:
- The value of \partial determines whether the two waves are in phase when they arrive at point P.
- If \partial is either zero or some integral multiple of the wavelength, then the two waves are in phase at point P and constructive interference occurs.
- Therefore the condition for bright fringes or constructive interference at P is: \partial = d \sin \theta_{bright} = m \lambda, where m = 0, \pm 1, \pm 2, …
- The number m is called the order number. The central bright fringe at \theta = 0 is called the zeroth-order maximum. The first maximum on either side, where m = \pm 1, is called the first-order maximum, and so forth.
- Conditions for Dark Fringes (Destructive Interference) in the two-slit experiment:
- When \partial is an odd multiple of \lambda/2, the two waves arriving at point P are 180° out of phase and give rise to destructive interference.
- Therefore the condition for dark fringes or destructive interference at P is: \partial = d \sin \theta_{dark} = (m + \frac{1}{2}) \lambda, where m = 0, \pm 1, \pm 2, …
- If m = 0 in this equation, the path difference is \partial = \lambda/2, which is the condition for the location of the first dark fringe on either side of the central (bright) maximum. Likewise, if m = 1, the path difference is \partial = 3\lambda/2, which is the condition for the second dark fringe on each side, and so forth
- Example: Light with wavelength \lambda is incident on two narrow slits spaced d meters apart. An interference pattern is visible on a screen located L meters from the slits. The second dark fringe is observed to be x meters from the central maximum. The diagram below shows the path of the light rays extending from each slit to the second dark fringe. What is the path difference?
- KPI 10.1.7 Numbering systems for bright and dark fringes:
- The m = 0 fringe occurs at \theta = 0, which corresponds to the central fringe. In addition, positive values of m indicate fringes above the central bright fringe; negative values indicate fringes below the central bright fringe.
- Fringe Spacing in an interference pattern:
- The distance between any two consecutive bright fringes or two consecutive dark fringes is called fringe spacing. The fringe spacing or thickness of a dark fringe or a bright fringe is equal.
- In addition to our assumption that L >> d, we assume that d >> \lambda. Under these conditions, \theta is small and we can use the small angle approximation \sin \theta ≈ \tan \theta.
- From the diagram above, if light propagates at an angle \theta relative to the normal to the slits, it is displaced a linear distance y = L \tan \theta on the distant screen.
- \tan \theta = \frac{y}{L}
- Substituting for \sin \theta = \tan \theta = \frac{y}{L} in the equation for bright fringes; d \sin\theta_{bright} = m\lambda
- The positions of the bright fringes measured from the center are; y_{bright} = m \frac{\lambda L}{d}
- Substituting for \sin \theta = \tan \theta = \frac{y}{L} in the equation for dark fringes; d \sin\theta_{dark} = (m + \frac{1}{2}) \lambda
- The positions of the dark fringes measured from the center are; y_{dark} = (m + \frac{1}{2}) \frac{\lambda L}{d}
- Young’s double – slit experiment provides a method for measuring the wavelength of light.
- If white light was used instead of a monochromatic light the following will be seen:
- Every color produces its own interference pattern, and we see them superimposed.
- The central maximum would be white.
- The first maximum is a full spectrum with violet on the inside and red on the outside.
- The second maximum is also a full spectrum, with red in it overlapping with violet in the third maximum. At larger angles, the light soon starts mixing to white again.
- Example 1: Monochromatic light of wavelength 600 nm is directed toward a double slit and forms an interference pattern on the screen. The two slits are 0.080 mm apart, and the screen is 2.0 m away.
- a. How far is the third dark fringe located from the center of the screen?
- y_{dark} = (m + \frac{1}{2}) \frac{\lambda L}{d}, where m = 0, \pm 1, \pm 2, …
- Therefore, for the third dark fringe, m = 2
- y_{dark} = (2 + \frac{1}{2}) \frac{(600 × 10^{-9} m)(2.0 m)}{0.080 × 10^{-3} m} = 0.0375 m
- b. Calculate the distance between adjacent bright fringes.
- \Delta y = y{(m+1)} - ym = \frac{\lambda L}{d} (m + 1) - \frac{\lambda L}{d} m
- \Delta y = \frac{\lambda L}{d} = \frac{(600 × 10^{-9} m)(2.0 m)}{0.080 × 10^{-3} m} = 0.015 m
- c. Does a larger slit create a larger separation between interference fringes?
- No. The fringe separation dependents on the slit separation and not the slit width.
- Example 2: Suppose you pass light from a laser through two slits separated by 0.010 mm and find that the third bright line on a screen is formed at an angle of 11º relative to the incident beam.
- a. What is the wavelength of the light?
- d \sin \theta{bright} = m \lambda \implies \sin \theta{bright} = \frac{m \lambda}{d}
- \lambda = \frac{d \sin \theta}{m} = \frac{(0.010 × 10^{-3} m) \sin 11º}{3} = 6.4 × 10^{-7} m
- b. What is the highest-order constructive interference possible with the system?
- Interference patterns do not have an infinite number of lines, since there is a limit to how big m can be.
- m = \frac{d \sin \theta}{\lambda}
- Since the maximum value of \sin \theta = 1,
- m = \frac{(0.010 × 10^{-3} m)(1)}{6.4 × 10^{-7} m} ≈ 15.6
- Therefore, the largest integer m can be is 15, or m = 15.
- Example 3: If white light is used in Young’s double-slit experiment rather than monochromatic light, how does the interference pattern change?
- Every color produces its own interference pattern, and we see them superimposed.
- The central maximum is white.
- Each maxima is a full spectrum with violet on the inside and red on the outside. At larger angles, the light soon starts mixing to white again.
- KPI 10.1.8 The fringe pattern in a double-slit experiment changes when different parameters are varied. The changes that occur to the fringe pattern depends on the relationship below: y ∝ \frac{\lambda L}{d}
- Accordingly the following is worth noting:
- If the wavelength (color) of the light changes, the fringe spacing also change if other parameters are kept constant. Example, if red color light is changed to a blue light, the wavelength decreases and so the fringe spacing also decreases.
- If only the distance between the slits and the screen (L) increases, the fringe spacing also increases, and vice versa.
- If the slit separation decreases, the fringe spacing increases, and vice versa.
- Increasing the slit width (a) will not change the fringe spacing. It only increases the amount of light entering the slits and thus make the fringes more brighter.
- Example 1: What happens to the spacing between fringes in a two-slit interference pattern if the following changes are made keeping other variables constant in each case.
- a. Slit separation is increased.
- y = \frac{\lambda L}{d} m
- If the slit separation (d) increases, the fringe spacing decreases. Thus we get narrower fringes.
- b. Color light is switched from red to blue.
- Switching the light from red to blue means the wavelength (\lambda) of the light has decreased. Again from the equation, a decrease in wavelength causes the fringe spacing to decrease so the fringes are closer to each other.
- c. The monochromatic light is replaced by while light.
- Colored fringes are formed at the screen if monochromatic light is replaced by the white light.
- d. The width of the two slits is increased.
- Intensity of light emitted by the slits increases as the width of the slit is increased. Thus, more brighter fringes are formed at the screen on increasing the width of the slits.
- Example 2: Choose the correct answer for each of the following questions below.
- i. In a two-slit interference pattern projected on a screen, are the fringes equally spaced on the screen
- (c) only for small angles
- ii. If the distance between the slits is doubled in Young’s experiment, what happens to the width of the central maximum?
- iii. A Young’s double-slit experiment is performed with three different colors of light: red, green, and blue. Rank the colors by the distance between adjacent bright fringes, from smallest to largest.
- Example 3: Monochromatic light with wavelength \lambda passes through two narrow slits that are a distance d apart. The resulting interference pattern appears as a series of alternating bright and dark regions on a screen located a length L meters behind the slits. How could the experiment be altered so that the spacing between the fringes on the screen is decreased?
Topic 10: Wave Optics
Subtopic 10.2: Single Slit Diffraction
- KPI 10.2.1
- Diffraction is the bending of waves around corners of an obstacle or through a gap (aperture) into the region of geometrical shadow of the obstacle/aperture.
- During diffraction, the wavelength (distance between the wave fronts) does not change.
- The diffraction decreases if the gap (aperture) becomes larger.