Question

In YDSE, light of wavelength $\lambda $ is used, which emerges in phase from the two slits at a distance $d$ apart. A transparent sheet of thickness $t$, refractive index $\mu =1.17$, is being placed over one of the slits. Where does the central maxima of the interference will appear from the centre of the screen? (Find the value of y also?)

$\begin{align}
  & A.\dfrac{D\left( \mu -1 \right)t}{2d} \\
 & B.\dfrac{2D\left( \mu -1 \right)t}{d} \\
 & C.\dfrac{D\left( \mu +1 \right)t}{d} \\
 & D.\dfrac{D\left( \mu -1 \right)t}{d} \\
\end{align}$

Answer 1

- Hint: When two waves constructively interfere and a bright fringe is observed at the centre. This is known as the central bright fringe. If one of the slits is covered, the fringes vanishes and there is a uniform illumination on the screen.

Complete step-by-step solution
First of all let us discuss the young’s double slit experiment. In modern physics, the double-slit experiment is a demonstration in which light and matter can display properties of both classically described waves and particles. Moreover, it is displayed as the fundamental probabilistic nature of the quantum mechanical process. Wave interference is defined as the phenomenon that happens when two waves coincide while traveling along the similar medium.
Here due to the introduction of the sheet in front of one of the slits, there will be some shift in the interference occurring. This is known as lateral shift. It is given as,
$\overline{X}=\dfrac{\beta }{\lambda }\left( \mu -1 \right)t$
Where $\lambda $ the wavelength, $t$ the thickness, $d$ is the slit width and $\mu $ is the refractive index.
And also we know that,
$\beta =\dfrac{\lambda D}{d}$
Substituting this in the equation,
$\overline{X}=\dfrac{D}{d}\left( \mu -1 \right)t$
Hence the correct answer is option D.

Note: The interference of waves will result in the medium to take on a shape that causes the net effect of the two individual waves upon the particles of the medium. The uses of interference are calculations made over long distances with lasers. In this case, the lasers can be helpful to measure very small distances over a range of so many miles. This is accomplished by splitting up the laser beam and reflecting it back from various surfaces.