Question

When a plastic thin film of refractive index 1.45 is placed in the path of one of the interfering waves then the central fringe is displaced through a width of five fringes. The thickness of the film, if the wavelength of light is $5890\;{\rm{\dot A}}$, will be:
A. $6.554 \times {10^{ - 4}}\;{\rm{cm}}$
B. $6.554 \times {10^{ - 4}}\;{\rm{m}}$
C. $6.54 \times {10^{ - 4}}\;{\rm{cm}}$
D. $6.5 \times {10^{ - 4}}\;{\rm{cm}}$

Answer 1

Hint: For the calculation of the thickness of the film, use the expression of path difference and rearrange it according to the desired result. The expression of path difference is,
${X_o} = \dfrac{n}{\lambda }\left( {\mu - 1} \right)t$

Complete step by step answer:
Given:
The refractive index of the thin film is \[\mu = 1.45\].
The displacement of the central fringe is ${X_o} = 5 \times {\rm{fringe}}\;{\rm{width}}$.
The wavelength of the light is $\lambda = 5890\;{\rm{\dot A}}$.
Write the expression of the path difference and explain each term.
${X_o} = \dfrac{n}{\lambda }\left( {\mu - 1} \right)t$
Here, ${X_o}$ is the displacement of the central fringe, $n$ is the width of the fringe, $\mu $ is the refractive index, $\lambda $ is the wavelength and $t$ is the time.
We will rearrange the above equation to obtain the expression of thickness.
$t = \dfrac{{{X_o}\lambda }}{{n\left( {\mu - 1} \right)}}$
Substitute the values in the above equation.
Therefore, we get
$t = \dfrac{{5n \times 5890\;{\rm{\dot A}} \times \dfrac{{{{10}^{ - 10}}\;{\rm{m}}}}{{1\;{\rm{\dot A}}}}}}{{n\left( {1.45 - 1} \right)}}\\
\implies t = 6.544 \times {10^{ - 6}}\;{\rm{m}} \times \dfrac{{100\;{\rm{cm}}}}{{1\;{\rm{m}}}}\\
\implies t = 6.544 \times {10^{ - 4}}\;{\rm{cm}}
$

So, the correct answer is “Option A”.

Note:
In these types of questions, always remember the expression of path difference and rearrange this equation according to the demand of the question. Like here, in this question, we need to determine the thickness of the film, so we rearranged the path difference expression according to the need of the question. The calculation and solution become easy and clear when we rearrange the equation.