Physics - Wave Optics Question with Solution | TestHub

Optical path difference for the light transmitted through oil is

$$\Delta x=\overline{2}\mu t\cos r$$

For normal incidence But the interface between oil and glass will produce an additional path difference of $\frac{\lambda }{2}$. There fore effective path difference $\Delta \mathrm{x}=2\mu \mathrm{t}+\frac{\lambda }{2}$

For constructive interference in transmitted light $2\mu \mathrm{t}+\frac{\lambda }{2}=\mathrm{n}\lambda ,\mathrm{n}=1,2,\dots \dots \dots$ or $2\mu \mathrm{t}=(2\mathrm{n}-1)\frac{\lambda }{2}$

Or $\lambda =\frac{4\mu \mathrm{t}}{(2\mathrm{n}-1)}=\frac{4\times 1.25\times 1\times 10^{-6}}{2\mathrm{n}-1}=\frac{5\times 10^6}{(2\mathrm{n}-1)}$

For $\mathrm{n}=1,\lambda =5\times 10^{-6}\mathrm{m}=5000\mathrm{nm}$

For $\mathrm{n}=2,\lambda =\frac{5\times 10^{-6}}{3}\mathrm{m}=1666.67\mathrm{nm}$

For $\mathrm{n}=3,\lambda =\frac{5\times 10^{-6}}{5}\mathrm{m}=1000\mathrm{nm}$

For $\mathrm{n}=4,\lambda =\frac{5\times 10^{-6}}{7}\mathrm{m}=714.29\mathrm{nm}$

For $\mathrm{n}=5,\lambda =\frac{5\times 10^{-6}}{5}\mathrm{m}=555.55\mathrm{nm}$

For $\mathrm{n}=6,\lambda =\frac{5\times 10^{-6}}{11}\mathrm{m}=454.54\mathrm{nm}$

The wavelength which are strongly transmitted in visible range are: 714.29 nm , 555.55 nm and 454.54 nm .

$$r=0,\cos r\to 1\Delta x=2\mu t$$