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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.
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Solution
Given:-
Wavelength of light used,
\[\lambda = 400 \times {10}^{- 9} to 750 \times {10}^{- 9} m\]
Refractive index of oil, μoil, is 1.25 and that of glass, μg, 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.
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