College physics optics questions, ask the masters.

1.After the white light passes through the first polarizer, the linear white light remains, and then through the second polarizer that changes direction, the linear light is further weakened. In this case, you can imagine yourself looking at the opposite side through two cracks.

2。This is similar to a linear white light passing through a prism.

Actually, these are all in books.

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Problem description:1After monochromatic light enters the crystal, it is birefringent due to the difference in the speed (wavelength and frequency change) of O and E light, so the energy of the monochromatic light changes, and the energy decreases? Or does it stay the same (one get one minus)?

2.In the Yangsui interference experiment, we have many approximations in the formula that proves that the fringe distance is x=l*d. I'm thinking:

Such an approximation is macroscopic, but since the light wave we are studying is related to its wavelength, it should be microscopic. So for light waves with a wavelength of only a few hundred nanometers, does such an approximation still have to be true?

3.Since the width of the bright stripes is equal, are the width of the dark stripes equal? Is it the same width as the bright stripe? How exactly do you determine their width?

Analysis: Hehe, I'm studying optics, let me give it a try.

1.Energy is constant. The total energy before birefringence can be divided into E light energy and O light energy, and there is no energy loss (not counting reflection and absorption) in the process of refraction. Therefore, the total energy after birefringence is equal to the sum of the two, and remains unchanged from before.

2.Establish. This is approximated using Fresnel.

List the expressions of the optical path difference, use the Taylor formula for the part with the root number, and approximate the spring Nianqi through the high-order infinitesimal relationship (so troublesome, remember that the book "College Physics" is so simple, but "Physical Optics" says so)!

3.Light and dark stripes are the same width.

Fringe spacing e= d l

Because the stripes do not overlap, the width is equal to the spacing.

Conditions for constructive interference:

2 n e + 0 = k and k is an integer = 2 n e k

k is in the visible range, only take k = 2, get.

nm

1、b。When oblique incidence, the stripes on the screen will move up or down, and the distribution up and down will no longer be symmetrical. Assume that within the original.

It's 5 at the top and 5 at the bottom, but after moving, it becomes 3 at the top and 7 at the bottom, and the level has become higher. Note that this up and down refers to the top and bottom of the grade 0 plain lines.

2, "The refractive index increases, there is no need to consider the half-wave loss, 2*d*n half a wavelength, so d 3, there is no reflected light.

1b2 has no half-wave loss. (1, 2 agree with the landlord).

3 When a beam of self-light from the air to the glass brick is incident at the angle of the brus daote back, the reflected light is complete.

Answer: Polarized light, but the refracted light is partially polarized, and when the light of the refracted part is reflected on the other side, it can be seen from Bruster's law that the light of the reflected part (because the angle of incidence is also a Brewster angle) is a linearly polarized light perpendicular to the incident surface. So choose B

Yes. Usually we are accustomed to symmetrical double slits, and the screen that forms the interference fringes is parallel and far enough from the double slits, which is not actually necessary for the formation of interference fringes.

In this problem, at the first interference o on the ox axis, it is obvious that the two light sources are in phase and reach the brightest; It then darkens in the direction of the ox; Considering that the two light sources are asymmetrical on the ox axis, the first dark pattern is not an odd multiple of the half-wavelength difference between the two optical pathlengths, and the brightness is not zero when the two are in phase again, so the point with zero intensity should be calculated in combination with the intensity after light divergence.

The maximum angle of incidence of incident light from the outside world is 90 degrees, and when you want to look at it in the water, you will see a circle, and the edge of the circle is the scene after the refraction of the light with an incident angle of 90 degrees. You should have done a problem, that is, put a light under the water that can propagate in all directions, it looks like a round spot on the surface of the water, and the other light cannot be emitted due to the large angle of incidence, and the light you come over is similar when it spreads into the water.

Knowing the angle of incidence and the refractive index of the edges of the great circle 4 3, we can calculate their exit angles, and let sin =sin90° (4 3)=3 4 then the relationship between radius r and height is.

h=r·cotα

Estimate the radius r, and let the athlete's height be the raised arm.

Refer to the first question here.

then radius r = then. h=

Obviously, when the angle of incidence is the largest, the corresponding angle of refraction is also the largest (except for total reflections).

When the directional light passes through the diameter, it does not change direction.

The angle of incidence of the edge rays is 30 degrees (this is obtained from the knowledge of right triangles), and the value of the critical angle c from glass to air can be determined by sinc 1 n (root number 2) 2 , c = 45 degrees.

So this beam of parallel light will be completely transparent (without total reflection), and the law of refraction will give n*sin30 degrees 1*sin r, r is the maximum refractive angle (root number 2) * r, and r 45 degrees.