Physics - Wave Optics Question with Solution | TestHub

For a single slit diffraction pattern, the distance of the $n$-th secondary maximum from the central maximum is given by:

$y_n=(2n+1)\frac{\lambda D}{2a}$

For the second maximum, $n=2$:

$y_2=\frac{5\lambda D}{2a}$

Given $D=1.3$ m, $a=0.65$ mm $=0.65\times 10^{-3}$ m, and $y_2=2.6$ mm $=2.6\times 10^{-3}$ m. Substituting these values:

$2.6\times 10^{-3}=\frac{5\times \lambda \times 1.3}{2\times 0.65\times 10^{-3}}$

$2.6\times 10^{-3}=\frac{5\times \lambda \times 1.3}{1.3\times 10^{-3}}$

$2.6\times 10^{-3}=5\times \lambda \times 10^3$

$\lambda =\frac{2.6\times 10^{-6}}{5}$

$\lambda =0.52\times 10^{-6}$ m

$\lambda =5200\times 10^{-10}$ m $=5200$ Å

Answer: $5200\mathring{A}$