Question

In a single slit diffraction experiment, the first minima for $\lambda_1 = 660 nm$ coincides with the first maxima for wavelength $\lambda_2$. Calculate $\lambda_2$.
A. 220 nm
B. 330 nm
C. 440 nm
D. 550 nm

Answer 1

Diffraction is a phenomenon by the virtue of which a light beam is split or bent when the wave encounters an obstacle or opening. In the single slit diffraction experiment, it is defined as the bending of waves around the corners of the slit through an aperture into the region of geometrical shadow of the aperture.

Formula used:
For first minima: $sin\theta = \dfrac{\lambda r}{d}$, For first maxima: $sin\theta = \dfrac{3\lambda\prime}{2d}$

Complete answer:
In this question, we are already given that the experiment is with a single slit. Also, given the two wavelengths, the position of first maxima and first minima of the two radiation coincides with one another. Thus, we just have to equate the location parameter of the two lights and we are done.
Now, we know that the location of both maxima and minima of two lights are at the same location and so the angle.
Thus by using formulas, For first minima: $sin\theta = \dfrac{\lambda r}{d}$, For first maxima: $sin\theta = \dfrac{3\lambda\prime}{2d}$, we get;
$sin\theta = \dfrac{\lambda r}{d} = \dfrac{3\lambda\prime}{2d}$
$\implies \lambda\prime = \dfrac{2\lambda}{3}$
Now, according to the question: $\lambda\prime = \lambda_2 \ and \ \lambda = \lambda_1$.
Thus $\lambda_2 = \dfrac23 \lambda_1 = \dfrac23 \times 660 = 440\ nm$
Thus, option C. is correct.

Additional information:
Diffraction is a phenomenon which is explained by the wave nature of light. After the split of a light beam, it undergoes interference and hence we are able to see a sustained pattern on the screen.

Note:
In interference or any wave related topic, a maxima is a point where the intensity of light is maximum after undergoing interference. Hence it corresponds to the brightest point in appearing on the screen. Similarly, minima is a point where the intensity of light is minimum after undergoing interference. Hence it corresponds to the dullest point in appearing on the screen.