Physics - Wave Optics Question with Solution | TestHub

Here, wavelength $\lambda =6000\mathring{A}=6000\times 10^{-10}\mathrm{m}$ Slit width $a=0.1\mathrm{mm}=0.1\times 10^{-3}\mathrm{m}$ In case of diffraction at single slit, the position of minima is given by $a\sin {\theta }_n=n\lambda$, where $n=1,2,3,\dots$. second minimum, $n=2\therefore a\sin {\theta }_2=2\lambda$ For small angle $\sin {\theta }_2\approx {\theta }_2$. $\therefore {\theta }_2=\frac{2\lambda }{a}$ Substituting the values, we get $\therefore {\theta }_2=\frac{2\times 6000\times 10^{-10}}{0.1\times 10^{-3}}=0.012$ radian