Distance between two coherent sources 1.1 mm

The distance between the two coherent sources is 1.1 mm, and the distance from the sources to the screen is 2.5 m. The sources emit light with a monochromatic wavelength of 0.55 μm. It is necessary to determine the number of interference fringes that fall per 1 cm of screen length.

To solve the problem, it is necessary to calculate the angle between the rays of light coming from the sources and hitting the screen. This angle can be calculated using the tangent of the rays. Thus, the angle between the rays of light is:

$$\theta = \tan^{ -1}\left(\frac{1.1}{2.5}\right) = 0.42 \text{ радиан}$$

To determine the number of interference fringes, you must use the formula:

$$m = \frac{d\sin\theta}{\lambda},$$

where $d$ is the distance between the sources, $\theta$ is the angle between the light rays, $\lambda$ is the wavelength of the light.

Using the values ​​given in the condition, we get:

$$m = \frac{1.1\cdot\sin(0.42)}{0.55\cdot10^{ -6}} \approx 1333$$

Thus, there are about 1333 interference fringes per 1 cm of screen length.

Distance between coherent sources

This digital product is an information product containing a description of the distance between two coherent light sources. The distance is 1.1 mm and is an important parameter for solving problems in the field of light interference.

This product will be useful for students, teachers and anyone interested in physics and optics. It is presented in digital format, which allows you to quickly and conveniently obtain the necessary information.

This product is an information product containing a description of the distance between two coherent light sources, which is 1.1 mm. This distance is an important parameter for solving problems in the field of light interference.

To solve a problem in which it is necessary to determine the number of interference fringes that fall per 1 cm of screen length, you can use the formula:

$m = \frac{d\sin\theta}{\lambda}$,

where $d$ is the distance between the sources, $\theta$ is the angle between the light rays, $\lambda$ is the wavelength of the light.

For this problem, the distance from the sources to the screen is 2.5 m, and the wavelength of light is 0.55 μm. The angle between the rays of light can be calculated using the tangent of the rays. Thus, the angle between the rays of light is:

$\theta = \tan^{ -1}\left(\frac{1.1}{2.5}\right) = 0.42 \text{ радиан}$

Substituting the values ​​into the formula, we get:

$m = \frac{1.1\cdot\sin(0.42)}{0.55\cdot10^{ -6}} \approx 1333$

Thus, there are about 1333 interference fringes per 1 cm of screen length.

This product can be useful for students, teachers and anyone interested in physics and optics. It is presented in digital format, which allows you to quickly and conveniently obtain the necessary information.

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This product describes an optical system consisting of two coherent light sources located 1.1 mm apart and a screen located 2.5 m from the sources. The sources emit monochromatic light with a wavelength of 0.55 µm.

The task is to determine the number of interference fringes that fall per 1 cm of screen length. To solve this problem, you can use the formula for calculating the number of interference fringes:

n = (d * sinθ) / λ,

where n is the number of interference fringes, d is the distance between the sources, θ is the angle between the light rays passing through each point on the screen, and λ is the wavelength of the light.

The angle θ can be calculated using the thin lens theorem:

θ = (λ * L) / (d * D),

where L is the distance from the sources to the screen, and D is the diameter of the hole through which the light rays pass.

By substituting the value of the wavelength, the distance between the sources and the distance to the screen, you can calculate the value of the angle θ. Then, using the value of the angle θ and the wavelength of the light, the number of interference fringes per 1 cm of screen length can be calculated.

I hope this description will help you understand what this product is and how to solve the problem associated with it. If you have any additional questions, feel free to ask me.

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