Physics - Wave Optics Question with Solution | TestHub

The shift in the central fringe due to the introduction of a transparent sheet of thickness $t$ and refractive index $\mu$ is given by:

$\Delta y=\frac{D}{d}(\mu -1)t$

The position of the $n^{th}$ bright fringe is given by:

$y_n=n\frac{D\lambda }{d}$

Given that the central fringe shifts to the position of the $30^{th}$ bright fringe, we have $\Delta y=y_{30}$:

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

$(\mu -1)t=30\lambda$

$\mu -1=\frac{30\lambda }{t}$

Substituting the given values $\lambda =6000\text{ A˚}=6000\times 10^{-8}\text{ cm}$ and $t=3.6\times 10^{-3}\text{ cm}$:

$\mu -1=\frac{30\times 6000\times 10^{-8}}{3.6\times 10^{-3}}$

$\mu -1=\frac{1.8\times 10^{-3}}{3.6\times 10^{-3}}$

$\mu -1=0.5$

$\mu =1.5$

Answer: $1.5$