Physics - Wave Optics Question with Solution | TestHub

Given, initial frequency of light, $$f_1=4\times 10^{14}{\mathrm{s}}^{-1}$$ Final frequency of light, $f_2=5\times 10^{14}{\mathrm{s}}^{-1}$ Change in wavelength $,\Delta \lambda ={\lambda }_1-{\lambda }_2$ or $$\Delta \lambda =\frac{c}{f_1}-\frac{c}{f_2}$$ $$\begin{array}{l}=\frac{3\times 10^8}{4\times 10^{14}}-\frac{3\times 10^8}{5\times 10^{14}}\\ =\frac{3\times 10^8}{10^{14}}\left(\frac{1}{4}-\frac{1}{5}\right)\\ =\frac{3\times 10^8}{10^{14}}\times \frac{1}{20}\\ =1.5\times 10^{-7}\mathrm{m}\end{array}$$ Now, we know that, Angular width of central maxima, $$\begin{array}{l}\theta =\frac{2\lambda }{d}\text{ or }\Delta \theta =\frac{2\Delta \lambda }{d}\\ d=\frac{2\Delta \lambda }{\Delta \theta }\end{array}$$ Here, $\Delta \theta =0.6$ radian $$d=\frac{2\times 1.5\times 10^{-7}}{0.6}=5\times 10^{-7}\mathrm{m}$$