Question

In Fraunhofer diffraction experiment at a single slit using a light of wavelength $400\text{nm}$, the first minimum is formed at an angle of ${30}^{\circ }$. The direction $\theta$ of the first secondary maximum is given by

• A. ${\sin }^{-1}\left(\frac{3}{8}\right)$
• B. ${\sin }^{-1}\left(\frac{3}{4}\right)$
• C. ${\sin }^{-1}\left(\frac{1}{4}\right)$
• D. ${\sin }^{-1}\left(\frac{2}{3}\right)$

Answer 1

The correct option is B ${\sin }^{-1}\left(\frac{3}{4}\right)$

The condition for diffraction minima in Fraunhofer diffraction is given by,

$b\sin \theta =n\lambda$

Where, $b-$ Width of the slit
$\lambda -$Wavelength of light used.

For first minima, $n=1$

$\therefore b\sin {30}^{\circ }=1\times 400\text{nm}$

$\Rightarrow b=800\text{nm}...\left(1\right)$

The condition for diffraction maxima in Fraunhofer diffraction is given by,

$b\sin \theta =(2n+1)\frac{\lambda }{2}$

For first secondary maxima, $n=1$

$\Rightarrow b\sin \theta =\frac{3\lambda }{2}$

Substituting the values of $b$ and $\lambda$

$\Rightarrow \left(800\text{nm}\right)\sin \theta =\frac{3}{2}\times \left(400\text{nm}\right)$

$\Rightarrow \sin \theta =\frac{3}{2}\times \frac{1}{2}$

$\Rightarrow \theta ={\sin }^{-1}\left(\frac{3}{4}\right)$

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

$\begin{array}{c}\text{Why this question?}\\ \text{Key Concept: For}{n}^{th}\text{diffraction minima}\\ b\sin \theta =n\lambda \\ \text{For}{n}^{th}\text{diffraction maxima}\\ b\sin \theta =(2n+1)\frac{\lambda }{2}\end{array}$