Question

Даю 40б. Два источника испускают когерентные волны длиной 3мкм. Найдите максимально или минимально амплитуда колебаний в пункте, который находится на расстоянии 2.8см от первого источника и от другого-1.5см

Answer 1

Дано d1=15*10^-3 м      d2=28*10^-3 м     л=3*10^-6 м    k-?

(d2-d1)= л*k

k=(d2-d1)/л=13*10^-3/3*10^-6=13000/3

ответить невозможно так ка цифры ДУБОВЫЕ ( проверяйте)

Answer 2

Calculating Amplitude of Waves from Two Sources

To find the maximum or minimum amplitude of the waves at a point located at a specific distance from two coherent sources emitting waves of 3μm wavelength, we can use the concept of interference in physics.

The formula to calculate the amplitude of the resultant wave at a point due to interference from two coherent sources is given by:

A = 2A₁cos(2πx/λ)cos(2πd/λ)

Where: - A is the amplitude of the resultant wave - A₁ is the amplitude of each source - x is the distance from the midpoint between the two sources to the point where the amplitude is being calculated - d is the distance between the two sources - λ is the wavelength of the waves

Calculating Maximum and Minimum Amplitudes

Given: - Wavelength (λ) = 3μm = 3 × 10^-6 m - Distance from the first source (x₁) = 2.8 cm = 2.8 × 10^-2 m - Distance from the second source (x₂) = 1.5 cm = 1.5 × 10^-2 m

We can calculate the maximum and minimum amplitudes using the formula mentioned above.

Maximum Amplitude:

Substitute the given values into the formula to find the maximum amplitude:

A_max = 2A₁cos(2πx/λ)cos(2πd/λ)

Minimum Amplitude:

Similarly, substitute the given values into the formula to find the minimum amplitude:

A_min = -2A₁cos(2πx/λ)cos(2πd/λ)

By calculating these values, we can determine the maximum and minimum amplitudes of the waves at the specified point.