Question

A glass surface is coated by an oil film of uniform thickness 1.00 × 10⁻⁴ cm. The index of refraction of the oil is 1.25 and that of the glass is 1.50. Find the wavelengths of light in the visible region (400 nm − 750 nm) which are completely transmitted by the oil film under normal incidence.

Answer 1

Given:
Wavelength of light used, $\lambda =400\times {10}^{-9}\mathrm{to}750\times {10}^{-9}\mathrm{m}$
Refractive index of oil, μ_oil, is 1.25 and that of glass, μ_g, is 1.50.
The thickness of the oil film, $d=1\times {10}^{-4}\mathrm{cm}={10}^{-6}\mathrm{m},$
The condition for the wavelengths which can be completely transmitted through the oil film is given by
$$\begin{gathered} \mathrm{\lambda }=\frac{2\mathrm{\mu }d}{\left(n+{\displaystyle \frac{1}{2}}\right)} \\ =\frac{2\times {10}^{-6}\times \left(1.25\right)\times 2}{\left(2n+1\right)} \\ =\frac{5\times {10}^{-6}}{\left(2n+1\right)}\mathrm{m} \\ \Rightarrow \mathrm{\lambda }=\frac{5000}{\left(2n+1\right)}\mathrm{nm} \end{gathered}$$
Where n is an integer.
For wavelength to be in visible region i.e (400 nm to 750 nm)
When n = 3, we get,
$$\begin{gathered} \lambda =\frac{5000}{\left(2\times 3+1\right)} \\ =\frac{5000}{7}=714.3\mathrm{nm} \end{gathered}$$
When, n = 4, we get,
$$\begin{gathered} \lambda =\frac{5000}{\left(2\times 4+1\right)} \\ =\frac{5000}{9}=555.6\mathrm{nm} \end{gathered}$$
When, n = 5, we get,
$$\begin{gathered} \lambda =\frac{5000}{\left(2\times 5+1\right)} \\ =\frac{5000}{11}=454.5\mathrm{nm} \end{gathered}$$
Thus the wavelengths of light in the visible region (400 nm − 750 nm) which are completely transmitted by the oil film under normal incidence are 714 nm, 556 nm, 455 nm.