The second angular minima location for a single-slit diffraction having slit width of $1.5 \mu \mathrm { m }$ and wavelength of 500 nm is
A) 13º.
B) 42º.
C) 25º.
D) 36º.

Question

Quizplus · Accepted Answer

The location of angular minima in single-slit diffraction is given by the formula $\sin \theta = \frac{m\lambda}{a}$, where $m$ is the order of the minima (for the second minima, $m = 2$), $\lambda$ is the wavelength, and $a$ is the slit width. Plugging in the given values, $m = 2$, $\lambda = 500 \times 10^{-9} \, \mathrm{m}$, and $a = 1.5 \times 10^{-6} \, \mathrm{m}$, we find $\sin \theta = \frac{2 \times 500 \times 10^{-9}}{1.5 \times 10^{-6}} = \frac{1000}{1500} = \frac{2}{3}$. Solving for $\theta$ gives $\theta \approx 41.81^\circ$, which rounds to $42^\circ$, making option B correct.

The Second Angular Minima Location for a Single-Slit Diffraction Having 1.5μm1.5 \mu \mathrm { m }1.5μm

The Second Angular Minima Location for a Single-Slit Diffraction Having $1.5 \mu \mathrm { m }$