91Ó°ÊÓ

Diffraction occurs when waves bend around the edges of an obstacle or aperture. In a single slit experiment, light diffracts when it passes through a narrow slit and creates a pattern of bright and dark fringes on a screen.
The dark fringes or minima in the pattern represent points where destructive interference occurs, causing the light waves to cancel each other.
These minima appear at specific angles \(\theta\) and are determined by the formula:
\[ a \sin \theta = m\lambda \]
Here, \(a\) is the slit width, \( \theta \) is the angle of diffraction, \( m \) is the order of the minimum (1 for the first minimum, 2 for the second, and so on), and \( \lambda \) is the wavelength of the light.
Essentially, the formula tells us that the position of these dark fringes depends on the wavelength of the light and the width of the slit.

The relationship between different wavelengths in a diffraction pattern can be explored through the conditions of their minima. In our exercise, two wavelengths \( \lambda_a \) and \( \lambda_b \) are chosen such that the first minimum of \( \lambda_a \) coincides with the second minimum of \( \lambda_b. \) This means:
\[ a \sin \theta = 1\lambda_{a} = 2\lambda_{b} \]
We can derive the relationship between \( \lambda_a \) and \( \lambda_b \) by setting the equations equal to each other:
\[ 1\lambda_{a} = 2\lambda_{b} \]
Solving this gives us: \[ \lambda_a = 2\lambda_{b} \]
This shows that the wavelength \( \lambda_a \) is twice that of \( \lambda_b. \) In terms of diffraction, this means that \( \lambda_a \) would produce minima at angles where \( \lambda_b \) produces every second minimum.

The angular position of the dark fringes or minima in a diffraction pattern depends on the wavelength of the light and the order of the minima. Using the formula: \[ a \sin \theta = m\lambda \]
we see that for a given slit width \( a \), the angle \( \theta \) is directly related to the product of the order of the minimum \( m \) and the wavelength \( \lambda. \)
If two minima coincide, their corresponding angles of diffraction must be such that:
\[ \sin \theta_{m} = m\lambda_{a} / a \]
and
\[ \sin \theta_{n} = n\lambda_{b} / a \]
must hold true. For our exercise, substituting \lambda_{a} = 2\lambda_{b} and checking for other coinciding minima reveals that:
\[ 2m = n \]
This means for every integer \( m \), there exists an even integer \( n \) such that the minima of both wavelengths coincide.
By understanding these principles, we can predict the angles at which dark fringes will appear for various wavelengths in different diffraction setups.