Question

In a Young's experiment, one of the slit is covered with a transparent sheet of thickness $3.6\times {10}^{-3}$ cm due to which position of central fringe shifts to a position originally occupied by ${30}^{th}$ bright fringe. The refractive index of sheet, if $\lambda =6000\mathring{A}$, is:

• A. $1.5$
• B. $1.2$
• C. $1.3$
• D. $1.7$

Answer 1

The correct option is A $1.5$

We are considering Young's double slit experiment.

Given,

Wavelength of the monochromatic source $=\lambda =6000\dot{A}=6000\times {10}^{-10}m$

Thickness of the transparent sheet, $t=3.6\times {10}^{-3}cm$

Refractive index of the sheet, $\mu =?$

The position of the ${30}^{th}$ bright fringe is given by, $x_n=n\frac{\lambda D}{d}$

That is,

$x_{30}=30\frac{\lambda D}{d}$

Shift of the central fringe is, $x_0=\frac{D}{d}(\mu -1)t$

where,

$D$ is the distance between the screen and slit.

$d$ is the distance between the slits.

That is,

$30\frac{\lambda D}{d}=\frac{D}{d}(\mu -1)t$

$⟹(\mu -1)=\frac{30\lambda }{t}$

$=\frac{30\times (6000\times {10}^{-10})}{(3.6\times {10}^{-5})}=0.5$

Then,

$\mu =1+0.5=1.5$