1. 34. Waves versus Particles; Huygens’ Principle and Diffraction
HW# 7
Chap 32: Pb. 12, Pb.43, Pb.46
Chap 33 : Pb. 14, Pb. 27, Pb.29
Due Friday, April 20

2. 33-2 The Thin Lens Equation; Magnification
Example 33-2: Image formed by converging lens.
What are (a) the position, and (b) the size, of the image of a 7.6-cm-high leaf placed 1.00 m from a mm-focal-length camera lens?
Solution: The figure shows the appropriate ray diagram. The thin lens equation gives di = 5.26 cm; the magnification equation gives the size of the image to be cm. The signs tell us that the image is behind the lens and inverted.

3. 33-2 The Thin Lens Equation; Magnification
Example 33-3: Object close to converging lens.
An object is placed 10 cm from a 15-cm-focal-length converging lens. Determine the image position and size (a) analytically, and (b) using a ray diagram.
Figure An object placed within the focal point of a converging lens produces a virtual image. Example 33–3.
Solution: a. The thin lens equation gives di = -30 cm and m = 3.0. The image is virtual, enlarged, and upright.
b. See the figure.

4. Example 33-3 Diagram

5. Problem 14
14. (II) How far apart are an object and an image formed by an 85-cm-focal-length converging lens if the image is larger than the object and is real?

6. 33-3 Combinations of Lenses
In lens combinations, the image formed by the first lens becomes the object for the second lens (this is where object distances may be negative). The total magnification is the product of the magnification of each lens.

7. 33-3 Combinations of Lenses
Example 33-5: A two-lens system.
Two converging lenses, A and B, with focal lengths fA = 20.0 cm and fB = 25.0 cm, are placed 80.0 cm apart. An object is placed 60.0 cm in front of the first lens. Determine (a) the position, and (b) the magnification, of the final image formed by the combination of the two lenses.
Figure Two lenses, A and B, used in combination. The small numbers refer to the easily drawn rays.
Solution: a. Using the lens equation we find the image for the first lens to be 30.0 cm in back of that lens. This becomes the object for the second lens - it is a real object located 50.0 cm away. Using the lens equation again we find the final image is 50 cm behind the second lens.
b. The magnification is the product of the magnifications of the two lenses: The image is half the size of the object and upright.

8. 33-4 Lensmaker’s Equation
This useful equation relates the radii of curvature of the two lens surfaces, and the index of refraction, to the focal length:
Figure Diagram of a ray passing through a lens for derivation of the lensmaker’s equation.

9. 33-4 Lensmaker’s Equation
Example 33-7: Calculating f for a converging lens.
A convex meniscus lens is made from glass with n = The radius of curvature of the convex surface is 22.4 cm and that of the concave surface is 46.2 cm. (a) What is the focal length? (b) Where will the image be for an object 2.00 m away?
Solution: Using the lensmaker’s equation gives f = 87 cm. Then the image distance can be found: di = 1.54 m.

10. 34 Waves versus Particles; Huygens’ Principle and Diffraction

11. 34-1 Waves versus Particles; Huygens’ Principle and Diffraction
Huygens’ principle: every point on a wave front acts as a point source; the wave front as it develops is tangent to all the wavelets.
Figure Huygens’ principle, used to determine wave front CD when wave front AB is given.

12. 34-1 Waves versus Particles; Huygens’ Principle and Diffraction
Huygens’ principle is consistent with diffraction: The wave bend in behind an obstacle, this is called diffraction
Figure Huygens’ principle is consistent with diffraction (a) around the edge of an obstacle, (b) through a large hole, (c) through a small hole whose size is on the order of the wavelength of the wave.

13. 34-2 Huygens’ Principle and the Law of Refractionθ1 θ2
Figure Refraction explained, using Huygens’ principle. Wave fronts are perpendicular to the rays.

14. 34-2 Huygens’ Principle and the Law of Refraction
Newton’s particle theory predicted that v2>v1 which is wrong when using geometry and speed of light speed changing in a medium.
Foucault confirmed the Huygen’s wave theory in 1850
Huygens’ principle can also explain the law of refraction (Snell’s Law).
As the wavelets propagate from each point, they propagate more slowly in the medium of higher index of refraction.
This leads to a bend in the wave front and therefore in the ray.

15. 34-2 Huygens’ Principle and the Law of Refraction
The frequency of the light does not change, but the wavelength does as it travels into a new medium:

16. 34-3 Interference – Young’s Double-Slit Experiment
If light is a wave, interference effects will be seen, where one part of a wave front can interact with another part.
One way to study this is to do a double-slit experiment:
Figure (a) Young’s double-slit experiment. (b) If light consists of particles, we would expect to see two bright lines on the screen behind the slits. (c) In fact, many lines are observed. The slits and their separation need to be very thin.

17. 34-3 Interference – Young’s Double-Slit Experiment
Monochromatic wavelength (single wavelength) through two slits.
If light is a wave, there should be an interference pattern.
Figure If light is a wave, light passing through one of two slits should interfere with light passing through the other slit.

18. Principle of Superposition
When waves overlap, their amplitudes add
Constructive
Destructive l + +
Phase
Shifted
180°, ,
or /2 l = =
Peaks line up with peaks
Peaks line up with troughs

19. What are PHASE differences?
Phase difference is a term used to describe the relative alignment of the peaks
IN PHASE= two waves are lined up—constructive interference
OUT of PHASE= one wave is upside down relative to the other—destructive interference

20. What causes Phase differences?
Light travels different distances just like any other wave d1
Source 1 d2
Source 2
Constructive Interference when in phase
Path difference is an integral number of wavelengths
∆d = d1-d2 = m
Where m is an integer
Destructive Interference when out of phase
Path difference is a half integral number of wavelengths
∆d = d1-d2 = (m+1/2)
Where m is an integer