Question

Light of wavelength $5000\mathring{A}$ is incident normally on a slit of width $2.5\times {10}^{-4}cm$. The angular position of second minimum from the central maximum is

• A. ${\sin }^{-1}\left(\frac{1}{5}\right)$
• B. ${\sin }^{-1}\left(\frac{2}{5}\right)$
• C. $\left(\frac{\pi }{3}\right)$
• D. $\left(\frac{\pi }{6}\right)$
• E. $\left(\frac{\pi }{4}\right)$

Answer 1

The correct option is B ${\sin }^{-1}\left(\frac{2}{5}\right)$

We know that,
Angular width of central maximum $=\frac{2\lambda }{a}$
where, $\lambda$ =wavelength of light,
a =width of single slit,
So, $sin\theta =\frac{2\lambda }{a}$
where $\lambda =5000\mathring{A}=5000\times {10}^{-10}m$
$a=2.5\times {10}^{-6}m$$\Rightarrow sin\theta =\frac{2\times 5000\times {10}^{-10}}{2.5\times {10}^{-6}}$
$=sin\theta =\frac{10000\times {10}^{-10}}{2.5\times {10}^{-6}}$

$=sin\theta =\frac{10}{25}$

$=sin\theta =\frac{2}{5}$

$\theta =si{n}^{-1}\left(\frac{2}{5}\right)$