Question

In Young’s double slit experiment, a 5th dark triangle is obtained at a point. If a thin transparent film is placed in the path of one of the waves, then the 7th bright fringe is obtained at the same point. The thickness of the film in terms of wavelength $\lambda $and refractive index $\mu $ will be
A. $\dfrac{{1.5\lambda }}{{\left( {\mu - 1} \right)}}$
B. $1.5\left( {\mu - 1} \right)\lambda $
C. $2.5\left( {\mu - 1} \right)\lambda $
D. $\dfrac{{2.5\lambda }}{{\left( {\mu - 1} \right)}}$

Answer 1

Hint:To solve this problem, we need to use the formulas for the position of dark and bright fringes in Young’s double slit experiment. Then by using them, first we will find the position of the 5th dark fringe and then that of the 7th bright fringe in terms of wavelength $\lambda $and refractive index $\mu $. Finally, we will find our required answer by considering the shift caused by the transparent film.

Formulas used:
${y_d} = \dfrac{{\left( {2n - 1} \right)\lambda D}}{{2d}}$, where, ${y_d}$ is the position of the dark fringe, $n$ is the number of the fringe, $D$ is the screen distance, $d$is the slit separation and $\lambda $is the wavelength
${y_b} = \dfrac{{n\lambda D}}{d}$, where, ${y_b}$ is the position of the bright fringe, $n$ is the number of the fringe, $D$ is the screen distance, $d$is the slit separation and $\lambda $is the wavelength and
$\Delta y = \dfrac{{\left( {\mu - 1} \right)tD}}{d}$, where, $\Delta y$ is the shift due to transparent film, $\mu $ is the refractive index, $t$ is the thickness of the film, $D$ is the screen distance and $d$ is the slit separation

Complete step by step answer:
It is given that In Young’s double slit experiment, the 5th dark triangle is obtained at a point.We know that the position of the dark fringe is given by ${y_d} = \dfrac{{\left( {2n - 1} \right)\lambda D}}{{2d}}$. Therefore, the position of 5th dark fringe is,
${y_d} = \dfrac{{\left( {2 \times 5 - 1} \right)\lambda D}}{{2d}} \\
\Rightarrow{y_d} = \dfrac{{9\lambda D}}{{2d}}$
Similarly, by using the formula for position of the bright fringe, ${y_b} = \dfrac{{n\lambda D}}{d}$, the position of the 7th bright fringe is given by,
$ {y_b} = \dfrac{{7\lambda D}}{d}$
If a thin transparent film is placed in the path of one of the waves, then the 7th bright fringe is obtained at the same point as the 5th dark fringe. This transparent film causes shift $\Delta y = \dfrac{{\left( {\mu - 1} \right)tD}}{d}$
$
\Rightarrow \dfrac{{9\lambda D}}{{2d}} + \dfrac{{\left( {\mu - 1} \right)tD}}{d} = \dfrac{{7\lambda D}}{d} \\
\Rightarrow \dfrac{{9\lambda }}{2} + \left( {\mu - 1} \right)t = 7\lambda \\
\Rightarrow \left( {\mu - 1} \right)t = 7\lambda - \dfrac{{9\lambda }}{2} \\
\Rightarrow \left( {\mu - 1} \right)t = \dfrac{{5\lambda }}{2} \\
\Rightarrow \left( {\mu - 1} \right)t = 2.5\lambda \\
\therefore t = \dfrac{{2.5\lambda }}{{\left( {\mu - 1} \right)}}$
Thus, the thickness of the film is $\dfrac{{2.5\lambda }}{{\left( {\mu - 1} \right)}}$.

Hence, option D is the right answer.

Note:In solving this problem, we have considered the formulas of positions of dark and bright fringes in Young’s double slit experiment. It is important to consider the shift due to the transparent film and add that to the position of the former fringe. As we have seen, the thickness of the film will be directly proportional to the wavelength and inversely proportional to the refractive index.