Скачать или смотреть In young`s experiment, the third bright band for the wavelength of light source A having wavelength

In young`s experiment, the third bright band for the wavelength of light source A having wavelength 6000Å coincides with the fourth bright band for source B in the same arrangement. Wavelength of light emitted by source B is
(a) 6289 Å (b) 4500 Å
(c) 2250 Å (d) 6000 Å
*Description:*

In Young's double-slit experiment, light from two different sources is used to create an interference pattern of bright and dark bands. The problem involves two sources, A and B, with different wavelengths. It states that the *third bright band* for *source A* with a wavelength of *6000 Å* coincides with the *fourth bright band* for **source B**.

*Key Concept:*
**Young's Experiment and Interference**: The interference pattern formed in Young's double-slit experiment depends on the wavelength of light. The bright bands (constructive interference) occur at positions where the path difference is an integer multiple of the wavelength.

**Formula for Bright Bands**:
The condition for constructive interference (bright bands) is:
\[
y_n = \frac{n \cdot \lambda \cdot D}{d}
\]
where:
\( y_n \) is the position of the \(n\)-th bright band,
\( \lambda \) is the wavelength of the light,
\( D \) is the distance between the screen and the slits,
\( d \) is the distance between the slits,
\( n \) is the order of the bright band (1st, 2nd, etc.).

Given that the third bright band for source A coincides with the fourth bright band for source B, the ratio of their wavelengths can be determined based on the position of the bright bands.

*Steps for Calculation:*
Let \( \lambda_A = 6000 \, \text{Å} \) (wavelength of source A).
The position of the third bright band for source A is proportional to \( 3 \cdot \lambda_A \).
The position of the fourth bright band for source B is proportional to \( 4 \cdot \lambda_B \), where \( \lambda_B \) is the unknown wavelength of source B.

Since both bands coincide at the same position, we have:
\[
3 \cdot \lambda_A = 4 \cdot \lambda_B
\]
Substitute \( \lambda_A = 6000 \, \text{Å} \):
\[
3 \cdot 6000 = 4 \cdot \lambda_B
\]
Solving for \( \lambda_B \):
\[
\lambda_B = \frac{3 \cdot 6000}{4} = 4500 \, \text{Å}
\]

**Answer**: The wavelength of light emitted by source B is **4500 Å**.

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