| Differences between revisions 3 and 10 (spanning 7 versions) | Back to page |
|
Size: 845
Comment:
|
← Revision 10 as of 11:04AM, Mar 06, 2014 ⇥
Size: 1840
Comment:
|
| Deletions are marked like this. | Additions are marked like this. |
| Line 1: | Line 1: |
| == Chapters 34 (Young's Double Slit) and Section 35-11 (Polarization) == | == Lecture 14: Chapters 34 (Young's Double Slit) and Section 35-11 (Polarization) == |
| Line 16: | Line 16: |
| Ans: C. Due to Snell's law, the fish looks higher than it really is. |
|
| Line 28: | Line 30: |
| Ans: A. |
|
| Line 31: | Line 35: |
| <<lia(superposition.png, align = center, width = 70 %)>> | <<lia(superposition.png, align = center, width = 50%)>> |
| Line 33: | Line 37: |
Ans: 3. |
|
| Line 46: | Line 52: |
Ans: B. One wavelength corresponds to $2\pi$. Quarter wave length difference here. == Young's double slit experiment == An interference pattern is seen from two slits. Now, cover one slit with glass, introducing a phase difference of $\pi$ (180$^\circ$; half wavelength) at the slits. How is the pattern altered? <<c>> <<lia(double_slit.png, align = center)>> <<c(1)>> A. pattern vanishes. A. pattern expands. A. bright and dark spots are interchanged. A. no change at all. Ans: C. Oveall phase difference is what matters. == Young's double slit experiment == In a double slit experiment, if the wavelength of the light is increased, the interference pattern will A. spread out. A. shrinks together. A. stays the same. A. disappears. Ans: A. $d\sin \theta = m \lambda$. With $d,m$ fixed, bigger $\lambda$ means bigger $\theta$. |
Lecture 14: Chapters 34 (Young’s Double Slit) and Section 35-11 (Polarization)
Gone fishing, 1
To shoot a fish with a gun, how should you aim?
- aim directly at the image
- aim slightly above
- aim slightly below
Ans: C. Due to Snell's law, the fish looks higher than it really is.
Gone fishing, 2
To shoot a fish with a laser gun, how should you aim?
- aim directly at the image
- aim slightly above
- aim slightly below
Ans: A.
Superposition
Ans: 3.
Phase difference
The two waves shown are
out of phase by $\pi$.
out of phase by $\pi / 2$.
out of phase by $\pi / 4$.
- in phase.
Ans: B. One wavelength corresponds to $2\pi$. Quarter wave length difference here.
Young’s double slit experiment
An interference pattern is seen from two slits. Now, cover one slit with glass, introducing a phase difference of $\pi$ (180$^\circ$; half wavelength) at the slits. How is the pattern altered?
- pattern vanishes.
- pattern expands.
- bright and dark spots are interchanged.
- no change at all.
Ans: C. Oveall phase difference is what matters.
Young’s double slit experiment
In a double slit experiment, if the wavelength of the light is increased, the interference pattern will
- spread out.
- shrinks together.
- stays the same.
- disappears.
Ans: A. $d\sin \theta = m \lambda$. With $d,m$ fixed, bigger $\lambda$ means bigger $\theta$.
Physics 5B UCSC




