College Physics Chapter 10 Optical Problems. How to tell if the interference is stronger or weaker

As shown in the figure below, S2 and S3 are two coherent light sources (with the same frequency and phase), and the optical path difference from this coherent light source to point P is: DSIN +R1-R2.

The refraction is how wide it is, and the reflection is how many litters of Miga.

In the air case: the distance from the 5th level of bright stripe to the center bright stripe is x 5*l*into d, l is the distance from the double slit to the light screen, and d is the distance between the double slits.

The optical path difference is δ 5 * into (because it is the 5th level of bright fringes, there is δ k* into the bright fringes) When the device is placed in a liquid, the wavelength of light is 1 into 1, there is x 7 * l* into 1 d, obviously there is 1 5 into 7

Since the frequency of light is constant, c into v into 1 and the refractive index n c v is obtained from the formula for wave velocity

The refractive index obtained is n into 1 7 5

1.The refractive index within the interlayer is not between the lens and the glass.

Between the refractive index of the glass plate, at the contact point between the convex surface of the lens and the glass, the thickness of the air layer is 0, and the optical path difference between the two reflected light is 2, so the center of the Newton's ring is a dark spot in the direction of the reflected light. The direction of the transmitted light is opposite to the fringe of the reflected light, so the center of the transmitted light Newton's ring is a bright spot.

If the refractive index in the sandwich is exactly between the refractive index of the lens and the glass plate, the center of the Newtonian ring of the reflected light is the bright spot, and the Newtonian ring of the transmitted light is the dark spot.

2.Of course, it does. The length of the rotation should be smaller than the diameter, and the area of the dark ring after you do this, the circumference or something, as long as the data related to the diameter is calculated, it will become smaller.

3.Because the rate of change of the optical path difference near the center is large, the horizontal distance required to change the optical path difference of one wavelength is smaller, that is, dense, because it is dense and so thin.

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In thin film interference, when the thickness of the film is an integer multiple of the half-wavelength of the light in the film, the reflected light interference is strengthened, the transmitted light is weakened, the number of photons of the transmitted light is reduced, and the projected light interference is weakened. The total number of photons of reflected light and transmitted light remains unchanged.

Due to the fluctuating nature of light, the reflected light at the two interfaces may interfere with each other (increase in intensity) or interfere with destructive (decrease in intensity), depending on their phase relationship. The phase relationship depends on the different pathlengths of the two reflected lights.

The optical path length depends on the film thickness, optical constant, and wavelength.

Wane. Because energy is conserved.

At this time, the optical path difference increases t(, and the fringes move by 5, indicating that the interference level has moved by 5 levels, that is, the change of the optical path difference makes the interference of the principle 0 level meet the interference of the 5th level, so the optical path difference should be 5 wavelengths.

So so, t=8 m=8*10 3nm.