High School Physics Optics Questions, Physics Optics Questions High School

This year's escort student will solve the problem for you:

On the one hand, the maximum range to see the luminous point is 10 meters.

Illustrate that the radius of the circle where the light can shoot out of the water is 5 meters.

On the other hand, the critical condition of the water surface is the total reflection condition (the exit angle is 90 degrees), according to the law of refraction: n*sina = 1 * sin90 and the sina = 1 n = 3 4

where a represents the angle of incidence at total reflection.

and by the geometric relation tana=5 h

The solution yields depth h=5*root number 7 3

Is 35 the original question? How do you feel that there are few conditions He swims back and forth in a certain direction, and the maximum range that can be seen in the luminous point s is 10m, which is the radius of the circle in which he glows.

The effect of surface tension will shrink the liquid surface, so that the cotton thread is stretched into a smooth arc, and the BC option is not smooth. The liquid in the question is an infiltrating liquid for the metal frame, so the arc formed is contracted.

A simple way to do this is to turn all the options upside down, and it is correct to look at the shape of the liquid level similar to the liquid level when the water is filled in the graduated cylinder.

You can't see it, so I suggest you try another chain.

If you connect o and ac, you can calculate the angle a0c as 75 degrees with a deflection angle of 45 and an incident angle of 30.

The angle of refraction of a can be analyzed as, n=

The corresponding AOD is 78 degrees, and the deflection angle of b can be analyzed to be 42 degrees, the refractive angle is 51°, and n=sin51 sin30

sinc=1 n=sin30 sin51=1 2sin51° options are clear.

I don't have a camera, I can't draw

What is the answer? How do I feel like A is right, B is not right C, and what is the D option that you have scribbled out, I don't understand.

Here's how I did it. (Draw a picture on paper yourself to see if it's right).

Set the angle OAC to the angle A. Then the angle oca is also equal to the angle a.

At point a, the angle of incidence of light A is equal to 30°, and the angle of refraction is the angle OAC, which is also equal to the angle of a, so the equation 1 is obtained according to the law of refraction: Na*Sin30°=1*Sina (where Na is the refractive index of the medium to the light of a color, 1 is the refractive index of air, and Sina is the sin value of the angle of A).

At point C, the angle of incidence is the angle OCA, which is also equal to the angle of a, and the angle of deflection is 45°, so the angle of refraction is a-45°, and the equation 2: 1*sina = na*sin(a-45°) is obtained according to the law of refraction.

Synapsis 1 and 2 are easy to get angles a = 75°. Bringing a=75° into Eq. 1 gives na=2*sin75°, so option b is incorrect.

From the above, the angle OAC is equal to A=75°.

Then we look at the B light, and at point A, the refractive angle of the B light is the angle OAD. From the title we know that the angle doc is equal to 3°, so the angle diac is equal to. So the angle oad = angle oac - angle diac = 75°.

That is, the angle of refraction of light b at point A is. The angle of incidence of B light at point A is 30°, so the deflection angle of B light is, so option A is correct.

As for the C option, we first need to find the refractive index nb of the medium against b light. At point a, according to the law of refraction, there is equation 3: nb*sin30°=1*. From this we find nb=.

Then, according to the critical angle formula, there is sinc=1 nb=1 (2*where c is the critical angle). So option c is not right either.

Option A is right, option D is not clear at the end.

It is probably equivalent to the fact that this problem is more troublesome to calculate, and it is necessary to use the Pythagorean determination to understand a triangle and make auxiliary lines. I won't talk about how to solve triangles, it's all mathematical knowledge. I'm here to show you how to construct this triangle.

According to the principle that the exit angle is greater than the angle of incidence, you draw the two optical paths closest to a and b, and draw the two optical paths illuminated from the glass, and then these two lines you draw form a quadrilateral with the straight lines p and ac. You align this trapezoidal into two parts along C, straight up. Let's analyze the triangle on the left (when analyzing, you also need to make auxiliary lines, use the Pythagorean theorem, calculate the positive Xuanqi and divide it by 1 2 to get it.)

Hope it helps!

This is a problem with the Bohr model, and the increase in kinetic energy is like a fact that a low-orbit satellite is faster than a high-orbit satellite.

Now let's talk about the question of the energy of the hydrogen atom.

The energy of the hydrogen atom includes the kinetic energy of the electron (in fact, it should include the kinetic energy of the nucleus, but the nucleus is used as the reference frame, so the kinetic energy of the nucleus is zero, so I will not mention it), the electric potential energy of the system of the nucleus and the electron outside the nucleus.

The kinetic energy is definitely increased, but what about the electric potential energy? The nucleus is positively charged, and the negatively charged electrons go from the "high rail" with low potential to the "low rail" with high potential, and the electric potential energy of course decreases, but how does the two energies change together? High school physics can't be judged directly!

But the Bohr model tells us that the total energy of the hydrogen atom is e1 = megaelectron volts, which is divided by n squares at the nth order, and a negative number will of course become larger after this number, such as megaelectron volts.

So choose A. It's very wordy, mainly to make you understand, in fact, you can choose this answer as long as the Bohr model is calculated, I am a physics teacher, I wish you progress!

When an electron jumps from the outer orbital to the inner orbital, it radiates energy outward, so the energy decreases.

The smaller the orbital radius of the electron, the greater the linear velocity. So the electron kinetic energy increases. (just like when a celestial body runs).

The nucleus is positively charged and the electrons are negatively charged. The different charges attract each other, the distance decreases, and the system energy decreases.

The electrons move around the nucleus, and the energy level of the inner shell is high, which has a higher energy than the outer shell, and its energy is also the kinetic energy increase.

It can also be said that the inner orbit moves fast and has a higher kinetic energy than the outer layer, just like a satellite.

The analogous gravitational formula solves for velocity change. The energy of the photons emitted during the transition is reduced!

When the transition to a lower energy level emits photons, the total energy decreases, but the kinetic energy increases according to the radius inversely proportional to the square of the velocity.

From the law of refraction we get n=sin sin

sin = cd sd = cd (sc +cd )sin =cd sa=cd (ac +cd ) so n (ac +cd ) (sc +cd) (ac +cd )=n (sc +cd )ac =n (sc +cd )-cd = (4 3) (ac=

The scale at point A is:

Refractive index n=sinr sin ,r angle of incidence, angle of refraction.

In the figure: the angle of view is 180, i.e. r=90°, sinr=1sin =10 (10 2+

So the refractive index should be 1

Select D

Let me explain, the sphere is symmetrical, and we will only take one of the sections for analysis.

Suppose the ab in my drawing is the diameter of this section, and take any point b on the diameter', from the sine theorem, sin 4=oc'/r ,sinθ3=ob'r, ob again' > oc', so 3 must be greater than 4, i.e., the angle of incidence is greatest only when the ray rays shoot upwards perpendicular to the disk.

and change the luminous position b', when b'When coincident with b, the angle of incidence sin 1 = (d 2) r because d 2 >ob', so sin 1 is maximum, i.e., 1 maximum, that is, the angle of incidence of the ray is the largest when it is shot vertically upwards at the edge of the disk.

If total reflection does not occur, there is a relationship as follows, sin 1(d 2) r<1 n

r>nd/2= mm

The answer is at least millimeters.

The calculation method of this problem is to first calculate the critical angle sin a= 1 n=1 , so a= degrees.

From the diagram you can see that the maximum angle of incidence is the incident point A (or point B) to the highest point of the semicircle O, so tan a= , so r=