Physics - Wave Optics Question with Solution | TestHub

The position of the $n$-th secondary maximum in a single slit diffraction pattern is given by:

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

For the second maximum, $n=2$:

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

Rearranging for the wavelength $\lambda$:

$\lambda =\frac{2a{y}_2}{5D}$

Given values are:
$D=1.4$ m
$a=0.66$ mm $=0.66\times 10^{-3}$ m
$y_2=2.8$ mm $=2.8\times 10^{-3}$ m

Substituting the values into the formula:

$\lambda =\frac{2\times 0.66\times 10^{-3}\times 2.8\times 10^{-3}}{5\times 1.4}$

$\lambda =\frac{3.696\times 10^{-6}}{7}$

$\lambda =0.528\times 10^{-6}$ m

$\lambda =5280\times 10^{-10}$ m $=5280\text{A˚}$

Answer: $5280\text{A˚}$