A Cassegrain astronomical telescope uses two mirrors to form the image. The larger (concave) objective mirror has a focal length of 50.0 cm. A small convex secondary mirror is mounted 43.0 cm in front of the primary. Light is reflected from the secondary through a hole in the center of the primary, thereby forming a real image 8.00 cm behind the primary mirror. What is the radius of curvature of the secondary mirror? / Since rays are originating from infinity, the image appears at the focal length or at 50 cm, which is 7 cm beyond the 2nd mirror. So object for 2nd mirror is 7cm behind the 2nd mirror.
1 / f2 = 1/ do + 1 / di
1 / f2 = 1/-7 + 1 /(43+8)
f2 = -8.11 cm / Mirrors: R = -2f
R = -2(-8.11)
R = 16.2 cm
A farsighted person uses glasses with a refractive power of 3.4 diopters. The glasses are worn 2.5 cm from his eyes. What is this person’s near point, when not wearing glasses? What is required refractive power for contacts?
Key Assumption:
The near point is generally corrected to be at 25 cm from the eye (not lens).
1/f = 1/di + 1 / do
3.4 = 1/di + 1/(.25-.025)
di = -0.957 meters (from the lens)
N = 95.7cm + 2.5 cm
N = 98.2 cm / (b)
1/f = 1 / di + 1 / do
1/f = -1/.982 + 1/(.25)
1/f = +2.98 diopters
The relaxed eyes of a patient have a refractive power of 48.5 diopters. (a) Is this patient nearsighted or farsighted? (b) Since this patient is nearsighted, find the far point. (Treat the eye as a single-lens system, with the retina 2.40 cm from the lens.)
(a) near sighted people like to see clearly in the distance, thus calculate refractive power for very distant objects.
1/f = 1 / di + 1/do
1/f = 1/0.024 + 1/∞
1/f = +41.7 diopters
Since this is less then the given refractive power, we know this person is near sighted. / (b)
1/f = 1 / di + 1 /do
48.5 = 1/0.024 + 1/ do
do = 14.6 cm

28.2 Double Slit

A laser with wavelength d/8 is shining light on a double slit with slit separation 0.500 mm. This results in an interference pattern on a screen a distance L away from the slits. We wish to shine a second laser, with a different wavelength, through the same slits. What is the wavelength λ2 of the second laser that would place its second maximum at the same location as the fourth minimum of the first laser, if d = 0.500 mm?

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d sinθ = m λ
  • d: inverse of the grating constant in meters, in our case (non-SI units of mm)
  • where m is the 1st order maxima, or the 2nd order maxima, etc.
o  if you want out of phase (maxima matching to minima), then
o  m - ½ corresponds to the 1st minima, when m = 2 then 2 - ½ is the 2nd minima, etc.
  • we also know sinθ must be equal in both cases if they are the same distance.

4th Minima
d sinθ = (4 - ½) λ1 / 2nd Maxima
d sinθ = 2 λ2 / 2 λ2 = (4 - ½) λ1
2 λ2 = (4 - ½) d/8
λ2 = 0.109 mm
28.3 Interference from Reflection off a Soap Film
What is the thinnest soap (nsoap = 1.33) film that appears black when illuminated with light with a wavelength of 490 nm?
The light reflected off the top of the soap has a half-wave phase shift, while the light that reflects off the bottom has no phase shift.
v = c / n
v = f λ
c / nair = f λair
f = c / (nair λair)
we also know freq can’t change / c / nsoap = f λsoap
c / nsoap = c / (nair λair) λsoap
1 / nsoap = 1 / (1 λair) λsoap
λsoap = λair / nsoap
λsoap = 490nm / 1.33
λsoap = 368 nm
Since the light reflected off the top and bottom of the bubble are 180° out of phase, the thickness of the bubble must correspond to ½ λsoap
½ λsoap = 368/2
½ λsoap = 184 nm
28.5 Rayleigh's criterion
Resolving Power / If you can read the bottom row of your doctor's eye chart (λlight = 570 nm), your eye has a resolving power of one arcminute, equal to 1/60°. If this resolving power is diffraction-limited, to what effective diameter of your eye's optical system does this correspond?
sinθd = 1.22 λ/D
sin(1/60°) = 1.22 570x10-9 / D
D = 2.39 mm / (The factor 1.22 comes from the fact that we are using a circular aperture for the diffraction pattern instead of an infinite slit.)