Optical instruments
A simple magnifying glass is just a converging lens. For it to form a virtual, upright, magnified image of an object, where must the object be placed?
C, converging lens produces virtual, upright, magnified image when object is placed inside the focal length between lens and f
1. In a double‑slit experiment, increasing the wavelength of the light causes the fringe spacing to:
Decrease
Increase
Stay the same
Become random
increase because y is proportional to wavelength
In single‑slit diffraction, the central maximum becomes wider when:
slit width decreases
f= 8*10^14 Hz
work function= 3.2 eV
whats the max kinetic energy
0.11 eV
list all the quantum numbers with n=2
1l, 0l, -2m, -1m, 0m, 1m, 2m, 1/2s, -1/2s
A simple magnifier has a focal length of 5 cm Assuming the near point of the eye is 25 cm, what is the maximum angular magnification when the final image is at infinity
When image is at infinity magnification= 25cm (assumed near point)/f, thus 25/5=5
bright fringes occur when the path difference between two slits is...
m*wavelength
bonus: whats m? how will it be stated in the question?
a laser passes through a narrower slit, the central maximum on the screen becomes...
wider and dimmer
Light falling on a metal surface causes electrons to be emitted from the metal by the photoelectric effect. In a particular experiment, the frequency of the light and the temperature of the metal are held constant. As the intensity of the light is gradually decreased to zero
a. the number of electrons emitted remains constant until the intensity reaches a certain value, then drops abruptly to zero.
b. the number of electrons emitted gradually decreases to zero.
c. the maximum speed of the emitted electrons decreases steadily.
d. the stopping potential gradually decreases.
B bc lower intensity means fewer photons which means fewer emitted electrons
2 reasons why 1s2, 1p2, 2s2, 2p10, 3d4 is not valid electron configuration
1. 1p2 is wrong, bc l has to be less than n, so l= 1 is not allowed when n=1
2. 2p10 is wrong, p holds max 6 electrons so this violates pauli exclusion principle
what is the corrective lens for farsightedness and nearsightedness
converging lens (convex) for farsightedness
diverging lens (concave) for nearsightedness
when one slit is partially blocked, the interference pattern...
has lower contrast due to less visible fringes
The first minimum in single slit diffraction occurs when...
asintheta=lambda
In a photoelectric effect experiment, light of frequency f (above the threshold frequency) shines on a metal surface. The intensity of the light is kept constant, but the frequency is gradually increased. What happens?
a. The number of emitted electrons decreases, but their maximum speed stays the same.
b. The number of emitted electrons increases, but their maximum speed stays the same.
c. The maximum speed of the emitted electrons increases, but the number emitted per second stays approximately the same.
d. Both the number of emitted electrons and their maximum speed increase.
C bc Kmax= hf-phi so increasing f increases hf , thus electrons come out with more kinetic energy and a higher speed, the number of electrons stays the same since constant intensity means that the photon rate is constant
A hydrogen atom electron transitions from n = 4 to n = 2.
energy is released bc electrons moving down to a lower energy level emit a photon
why does a hyperopic person sees distant objects clearly but nearby objects appear blurry?
hyperopic= farsightedness
the eye is too short, so the image forms behind the retina
usually, in a normal eye the cornea and lens bend incoming light so that the image lands exactly on the retina, but with farsightendness the focal point ends up behidn the retina, which is physically impossible, and the light converges after it so the image is blurry
if the screen is moved farther away in a double slit experiment, the fringe spacing
increases
decreases
stays constant
disappears
increases
if the slit width is doubled the angular width of the central maximum
doubles
halves
stays the same
becomes zero
halves bc width is proportional to 1/a
wavelength= 400nm, stopping potential is 1V, find the work function of the metal in eV (if hc= 1240eV*nm)
E= hc/wavelength
Kmax= Ephoton - phi
Kmax= eV
E= 1240/400=3.10eV, K=eV=1 (since V was stated to be 1)
using K= E- phi: 3.1-phi =1 means that phi= 2.10eV
How many unique sets of quantum numbers (n, ℓ, mℓ, ms) are possible for electrons in the n = 3 shell, and how many orbitals does this correspond to?
For n = 3:
a person needs reading glasses, this means their near point is...
farther than normal, near point is the closest distance at which your eye can focus on a object clearly using max accomodation, for a normal aduly near point is 25cm, so if someone needs glasses then they cant focus on objects at 25cm (their near point might be 40-100cm ish) and need converging lens to shift image inward so it lands on the retina
Two waves interfere destructively when their phase difference is:
0
pi/2
pi
2pi
pi because pi = half cycle out of phase, so they cancel each other out
draw interference vs diffraction
draw constructive vs destructive
diffraction_and_interference.png (1917×1078)
constructive= waves merge together and get bigger
destructive= waves cancel out
diffraction is when light bends around obstacles if needed, interference is when waves interact with each other
Two monochromatic light sources, A and B, shine on the same metal surface in separate experiments.
Which statement is true?
a. Source A emits electrons with greater maximum kinetic energy than source B.
b. Source B emits more electrons per second than source A.
c. Both sources emit electrons, but source B emits more because it is more intense.
d. Neither source emits electrons because intensity must exceed a threshold.
A is correct bc electron emission depends on frequency not intensity, if A is above the threshold then A does emit electrons, since B is below the threshold it emits zero electrons no matter how bright the light is, source A having higher frequency means higher hf means greater kinetic energy
Bohr’s math for energy levels is right for hydrogen but his physical interpretation of how electrons move is wrong.
What’s wrong in Bohr’s model is the mechanical picture: electrons are not little particles in fixed circular orbits. In quantum mechanics, they are described by wavefunctions and orbitals, not orbits.
The Bohr model gets the wavelengths right because, even though its picture of electrons in circular orbits is wrong, its energy quantization assumption (where E= -13.6eV/n^2) is right for hydrogen.