Question

A plane wave front of light of wavelength $6000\mathring{\text{A}}$ is incident upon a slit of $0.2\text{mm}$ width, which enables Fraunhofer's diffraction pattern to be obtained on a screen placed $2\text{m}$ away from the slit. Width of the central maxima $\left(\text{in mm}\right)$ will be,

• A. $10$
• B. $12$
• C. $8$
• D. $2$

Answer 1

The correct option is B $12$

Width of the central maximum
$y_0=2\beta =\frac{2\lambda D}{d}$
$\lambda \to$ Wavelength of light
$D\to$ Distance between slit and the screen
$d\to$ Width of the slit
$y_0=\frac{2\times 6000\times {10}^{-10}\times 2}{0.2\times {10}^{-3}}=12\times {10}^{-3}m=12\text{mm}$

Hence, $\left(B\right)$ is the correct answer.

Why this question: –
$\begin{array}{c}\text{Why this question?}\\ \text{To understand the characteristics of diffraction pattern.}\end{array}$