Question

The first minima for the wavelength $\lambda_1 = 660 \, \text{nm}$ coincides with the first maxima of some other wavelength $\lambda_2$, in the single-slit diffraction experiment of light. Find out the value of wavelength $\lambda_2$.

Hint:

In single slit diffraction: minima $\propto m \lambda$, while maxima can overlap with minima of another wavelength.

Answer 1

Step 1: Condition for diffraction minima.
For single slit: \[ a \sin \theta = m \lambda_1, \quad m = 1, 2, 3, ... \] For first minima: \[ a \sin \theta = \lambda_1. \]
Step 2: Condition for diffraction maxima.
For interference maxima of another wavelength $\lambda_2$: \[ a \sin \theta = n \lambda_2, \quad n = 1, 2, 3, ... \] For first maxima: $n=2$: \[ a \sin \theta = 2 \lambda_2. \]
Step 3: Equating conditions.
Since first minima of $\lambda_1$ coincides with first secondary maximum of $\lambda_2$: \[ \lambda_1 = 2 \lambda_2. \] \[ \lambda_2 = \frac{\lambda_1}{2}. \]
Step 4: Substitution.
\[ \lambda_2 = \frac{660}{2} = 330 \, \text{nm}. \]
Step 5: Conclusion.
Hence, the required wavelength is $\lambda_2 = 330 \, \text{nm}$.