Question

In the fraunhaufer diffraction from a single slit illuminated by polychromatic light, the first minimum with wavelength \(\lambda_{1}\) is found to be coincident with the third minimum at \(\lambda_{2}\). Then the relationship between these two wavelengths is (1) \(3 \lambda_{1}=2 \lambda_{2}\) (2) \(3 \lambda_{1}=\lambda_{2}\) (3) \(\lambda_{1}=3 \lambda_{2}\) (4) \(2 \lambda_{1}=3 \lambda_{2}\)

Short answer

Option (3): \( \lambda_{1} = 3 \lambda_{2} \).

Step-by-step solution

Understanding Diffraction Minima

The position of minima in a diffraction pattern is given by the formula: \[ a \times \text{sin} \theta = n \times \frac{\lambda}{2} \] where \(a\) is the slit width, \( \theta \) is the diffraction angle, \( n \) is the order of the minima, and \( \lambda \) is the wavelength.

Identify Given Information

It's given that the first minimum of wavelength \( \lambda_1 \) coincides with the third minimum of wavelength \( \lambda_2 \). This means the diffraction angles are the same for these minima.

Set Up Equality for Minima

Using the condition for the minima positions, write the equations for the first and third minima: a.sin(\( \theta \)) = \(1 \times \lambda_1 \)For the third minimum: a.sin(\( \theta \)) = \(3 \times \lambda_2 \)

Equate the Two Diffraction Conditions

Since these two minima coincide: \(a \times \sin \theta = \lambda_1 = 3 \times \lambda_2 \)

Solve for the Relationship

By solving \( \lambda_1 = 3 \times \lambda_2 \), we simplify to: \( \lambda_{1} \) coincides with \(3 \lambda_{2} \)

Key concepts

Diffraction Minima

Diffraction minima are the dark spots in a diffraction pattern where destructive interference occurs. In the context of a single-slit diffraction setup, it is important to understand that these minima occur at specific angles depending on the wavelength of the light and the width of the slit.
The position of these minima is given by the formula: \[ a \times \text{sin} \theta = n \times \frac{u}{2} \] Here, \(\text{a}\) is the width of the slit, \(\theta\) is the diffraction angle, \(\text{n}\) is the order of the minima (1st, 2nd, 3rd, etc.), and \(u\) is the wavelength of the light. In simpler terms, this equation helps us determine where on a screen a dark fringe (where no light appears) will form when light passes through a narrow slit.

Polychromatic Light

Polychromatic light consists of multiple wavelengths. When polychromatic light passes through a single slit, each wavelength undergoes diffraction, creating a series of diffraction patterns that overlap with each other.
Now, since different wavelengths diffract at different angles, the patterns produced by each wavelength will not be identical. This results in a complex pattern where the diffraction minima for one wavelength will not directly correspond to those of other wavelengths except at specific points where they might coincide.

Wavelength Relationship

In this exercise, we are given that the first minimum for one wavelength (\(u_{\text{1}}\)) coincides with the third minimum for another wavelength (\(u_{\text{2}}\)). Using the formula for diffraction minima, setting up the equality for both wavelengths shows a direct relationship between them.
For the first minimum of \(ν_{\text{νufuna}}\) and the third minimum of \(ν_{\text{2}}\) we have: \[ a \times \text{sin} \theta = u_{\text{νufuna}} = 3 u_{\text{2}} \] This essentially means that: \( u_{\text{νufuna} = 3 u_{\text{2}} \).
So, the wavelength \(ν_{\text{νufuna}}\) must be three times the wavelength \(ν_{\text{νufuna}}\), leading us to the solution. Always remember that understanding these relationships allows us to predict and explain how light of different wavelengths will behave in diffraction experiments.