Question

In Young's double- slit experiment, the $y-$ coordinate of central maxima and ${10}^{th}$ maxima are $2\text{cm}$ and $5\text{cm}$ respectively. When the $\text{YDSE}$ apparatus is immersed in a liquid of refractive index $({\mu }_w=1.5)$, the corresponding $y-$ coordinate will be:

• A. $2\text{cm},7.5\text{cm}$
• B. $3\text{cm},6\text{cm}$
• C. $2\text{cm},4\text{cm}$
• D. $\frac{4}{3}\text{cm},\frac{10}{3}\text{cm}$

Answer 1

The correct option is C $2\text{cm},4\text{cm}$

We know that, fringe width, $\beta =\frac{\lambda D}{d}$

$\Rightarrow \beta \propto \lambda \propto \frac{1}{\mu }$

When $\text{YDSE}$ setup immersed in liquid,


The distance between central maxima and ${10}^{th}$ maxima is,

$\Delta y{}^{\prime }=y_{10}^{{}^{\prime }}-y_0^{{}^{\prime }}$

$\Rightarrow \Delta y{}^{\prime }=\frac{10{\lambda }_{w}D}{d}-\frac{{\lambda }_{w}D}{d}$

$\Rightarrow \Delta y{}^{\prime }=\frac{1}{{\mu }_w}[\frac{10{\lambda }_{a}D}{d}-\frac{{\lambda }_{a}D}{d}]$

$\Rightarrow \Delta y^{{}^{\prime }}=\frac{1}{{\mu }_w}\left[y_{10}-y_0\right]=\frac{\Delta y}{{\mu }_w}$

Thus, the distance between both maxima will reduce to,

$\Delta y{}^{\prime }=\frac{5-2}{\frac{3}{2}}=2\text{cm}$

As the $y-$ coordinate of central maxima doesn't depend upon wavelength, therefore it will be remained same as in air.

$\therefore y_0^{{}^{\prime }}=2\text{cm}$ and $y_{10}^{{}^{\prime }}=2+2=4\text{cm}$

Hence, $\left(\text{C}\right)$ is the correct answer.

Why this question ? Note: Position of central maxima will remain unaffected by the medium in which YDSE is performed.