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,

\[\lambda = 400 \times  {10}^{- 9}   to  750 \times  {10}^{- 9}   m\]

Refractive index of oil, μ_oil, is 1.25 and that of glass, μ_g, is 1.50.

The thickness of the oil film,

\[d = 1 \times {10}^{- 4} cm = {10}^{- 6} m,\]

The condition for the wavelengths which can be completely transmitted through the oil film is given by

\[\lambda = \frac{2\mu d}{\left( n + \frac{1}{2} \right)}\]

\[     = \frac{2 \times {10}^{- 6} \times \left( 1 . 25 \right) \times 2}{\left( 2n + 1 \right)}\]

\[     = \frac{5 \times {10}^{- 6}}{\left( 2n + 1 \right)}  m\]

\[ \Rightarrow   \lambda = \frac{5000}{\left( 2n + 1 \right)}  nm\]

Where n is an integer.

For wavelength to be in visible region i.e (400 nm to 750 nm)

When n = 3, we get,

\[\lambda = \frac{5000}{\left( 2 \times 3 + 1 \right)}\]

\[   = \frac{5000}{7} = 714 . 3  nm\]

When, n = 4, we get,

\[\lambda = \frac{5000}{\left( 2 \times 4 + 1 \right)}\]

\[     = \frac{5000}{9} = 555 . 6  nm\]

When, n = 5, we get,

\[\lambda = \frac{5000}{\left( 2 \times 5 + 1 \right)}\]

\[     = \frac{5000}{11} = 454 . 5  nm\]

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.