Physics - Wave Optics Question with Solution | TestHub

The condition for the $n$-th secondary maximum in a single slit diffraction pattern is given by $a\sin \theta =(n+\frac{1}{2})\lambda$.

For the second secondary maximum ($n=2$) of wavelength $\lambda$:

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

For the third secondary maximum ($n=3$) of wavelength ${\lambda }_0$:

$a\sin \theta =\frac{7}{2}{\lambda }_0$

Since the two maxima coincide, their angular positions are equal:

$\frac{5}{2}\lambda =\frac{7}{2}{\lambda }_0$

${\lambda }_0=\frac{5}{7}\lambda$

Substituting $\lambda =6195\text{ A˚}$:

${\lambda }_0=\frac{5}{7}\times 6195=5\times 885=4425\text{ A˚}$

Answer: $4425\mathring{A}$