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Light is reversible, whether it travels in a straight line or is reflected in the same medium or refracted in different media. In total reflection, as long as you have a large enough angle of incidence in a photophobic medium (close to 90 degrees), you will also find that the light is reversible. The first floor violates scientific common sense, and the third floor concludes correctly but the argument is completely unreasonable

Optical path reversibility is just a technical method of optics, especially linear optics, for the sake of ease of analysis. That is, in order to help understand, or when doing geometrical optical figure analysis, this method can be agreed with equivalently. For most geometrical optics problems, it can be considered that the law of reversibility of the optical path, or the problem-solving technique, is effective, and the critical angle you mentioned does not contradict this law.

The optical path reversible is based on two fundamental Newtonian principles of mechanical physics, namely the conservation of momentum and the conservation of energy. The theoretical basis of this law is Newton's theory of particles of light. And the critical angle problem you raised is also very normal, which is equivalent to a completely elastic collision.

All the abilities of the light particles are returned by **, and the complete reflection occurs, and no energy is transferred to the refracted light particles, resulting in no refraction.

In essence, modern optics has almost abandoned the method of light particles to explain linear optical problems such as the refraction and reflection of light. Using the wave theory of light, it is possible to solve the relevant problems in a clear and accurate way by adding proof in mathematical language.

As for the word "reversible", it is also a figurative statement. Spatially you can go back backwards and call it reversible? All in all, the optical path reversibility is only a technical means to deal with the problem, and it is not worth in-depth scrutiny. There is no mathematical foundation that is strictly forbidden in the first place.

Reversible! Generally speaking, light is emitted from air into other media, but when light is emitted from other media into air, the refracted rays are deflected in the direction away from the normal, and the angle of incidence is less than the angle of refraction!

Reversible, the critical angle can also be explained, if the critical angle is reached, the light cannot be emitted!

Can be reversed! Refraction is produced by the different velocities of light in different media! Since the medium has not changed, of course it can!

It is only reversible in geometric optics and paraxial optics, and the critical angle you are talking about is the total reflection angle, which can be calculated according to the law of refraction, so that the refractive angle is equal to 0.

The outgoing light cannot be absolutely horizontal, which is a matter of limits, so I feel irreversible.

Categories: Science & Engineering.

Problem description: There is a parallel fluorescent lamp installed on the ceiling, which is covered by a grid-like iron frame (composed of horizontal and vertical iron bars), looking at the ceiling from under the lamp, at a certain distance, you will find that the iron bars parallel to the fluorescent lamp cannot be seen, why?

Analysis: Mainly the reflection of light.

The back of the fluorescent tube is a white reflector (white iron sheet), and the fluorescent lamp is not a point light source, the light emitted is divergent, if it is a directional light, then there will be shadows, but the divergent light and the reflection of the reflector fill these shadows.

/2 / l = d / l

d = mm

Move in the direction of the ridge line town mausoleum;

The stripes at the wire move 9 stripes, 9 * 2 = mm, that is, the diameter of the wire increases and the number of mm is added. Imperial Rules.

This is caused by the reflection of light, because your child's image is projected, hugging the mirror, and then reflected back into your eyes, so you can see a lower image than the girl you meet.

Because your viewing angle is very high, you look down in the mirror, so you can see the baby who is lower than the mirror, and if the mirror is lower, you can see the ground.

Because what you see is a reflection of the mirror, and you can usually see that position.

There is light shining on the baby, and then the light shines on the mirror, and the reflection of the light enters his own eyes. (The incident light and the reflected light are at the same angle from the mirror).

Mirror imaging is formed by the reflection of light.

The reflection theorem states that the incident and reflected rays are located on either side of the normal, and the angle of reflection is equal to the angle of incidence.

Because the baby is lower than the mirror, the incident light from the baby should be tilted upwards to enter the mirror. That is, the angle of incidence is below the normal.

If your eyes are high, then the reflection angle coming out of the mirror is just above the normal. As long as the angle of reflection of the light entering your eyes is equal to the angle of incidence, you can see the baby's image.

If the eye is positioned low (so that the angle of reflection is less than the angle of incidence), the baby's image will not be visible.

It should be 4 approx. = 3And the "depth of the water layer" you ask should be the thickness of the water layer. Here's why:

Let d = 4cm, n1 =, n2 =, ether layer thickness (i.e. depth) d2, water layer depth d1

Thickness of the water layer = depth of the water layer - depth of the ether layer. Take a very small angle of incidence:

Approximately, sinx=tanx

Think of the image and the real thing in the same vertical line.

And then according to the law of refraction:

sin i sin r = n i.e. tan i tan r = n. If you draw a diagram to find the geometric relationship, it is easy to know that d2=d n2, and then find the geometric relationship, then let the horizontal distance between the three incident points be x1 and x2 from left to right

Same as above, there is sin i'/sin r'=n1 n2 i.e. tan i'/tan r'=n1 n2, again tan i'=x2/d, tan r'=x1 d, then x1 x2=n1 n2

The triangle in the upper left part is similar to d2 d1=x2 (x1+x2).

So the solution is d1=d(n1+n2) (n1n2)

The thickness of the water layer = the depth of the water layer - the depth of the ether layer, that is, "sought" = d1 - d2 = d n1 = 4

Still 4cm, analysis: perpendicular to the liquid level, the light is not refracted, so there is no parallax.

This teacher is probably not good at preparing for class, and the lectures in class are all about the scientific research he has participated in (or write ** to check the corresponding information), I can only give a general direction as an optical major, and then you can combine the content of the class (definitely not less, looking for classmates' notes) and search for relevant information on the Internet.

1. This should be done with a photoelectric theodolite (God.

5. Shen 6) for spatial positioning tracking.

2. There are many measurement methods for this question, give you a few plans, check the notes. Photoelectric encoders are used for precise measurement of angles, and interference devices can be used for flatness detection (I don't know which type your teacher uses, there are many models); Perpendicularity detection can be done with a laser collimator; Surface shape errors can only be done with an interfering device.

3. The lithography machine works in a timely manner, and it is available in one check.

4. There are many methods, and a code let the wheel belong to the optical processing technology, which can not be abbreviated for non-professionals.

Your old teacher is teaching these things to the elective students instead of getting started with literacy, and the old teachers who are not very old or just graduated are young people, waste your time and energy, deal with it. I can help you so much!