Question

Light of wavelength $589.3\text{nm}$ is incident normally on the slit of width $0.1\text{mm}$. What will be the angular width of the central maximum, obtained on a screen placed at a distance of $1\text{m}$ from the slit?

• A. ${0.67}^{\circ }$
• B. ${1.02}^{\circ }$
• C. ${1.34}^{\circ }$
• D. None of the above

Answer 1

The correct option is A ${0.67}^{\circ }$

Angular width of central maxima,

$=\frac{2\lambda }{d}=\frac{2\times 589.3\times {10}^{-9}}{0.1\times {10}^{-3}}\text{in rad}$

$=0.0117\text{rad}=0.0117\times \frac{180}{\pi }={0.67}^{\circ }$

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

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