Question

A soap film of thickness 0.0011 mm appears dark when seen by the reflected light of wavelength 580 nm. What is the index of refraction of the soap solution if it is known to be between 1.2 and 1.5?

Answer 1

Given:-

Thickness of soap film, d =0.0011 mm = 0.0011 × 10⁻³ m

Wavelength of light used,

\[\lambda = 580  nm = 580 \times  {10}^{- 9}   m\]

Let the index of refraction of the soap solution be μ.

The condition of minimum reflection of light is 2μd = nλ,

where n is an interger = 1 , 2 , 3 ...

\[\Rightarrow \mu = \frac{n\lambda}{2d} = \frac{2n\lambda}{4d}\]

\[           = \frac{580 \times {10}^{- 9} \times \left( 2n \right)}{4 \times 11 \times {10}^{- 7}}\]

\[           = \frac{5 . 8\left( 2n \right)}{44} = 0 . 132\left( 2n \right)\]

As per the question, μ has a value between 1.2 and 1.5.

So,

\[n = 5\]

So, \[\mu = 0 . 132 \times 10 = 1 . 32\]

Therefore, the index of refraction of the soap solution is 1.32.