Question

${}^{\prime }{n}^{\prime }$ transparent slabs of refractive index $1.5$ each having thickness $1\text{cm},2\text{cm},...n\text{cm}$ are arranged one over the other. A point object is seen through this combination from top with perpendicular light. If the shift of the object by combination is $5\text{cm}$. Then the value of ${}^{\prime }{n}^{\prime }$ is

Answer 1

For a small angle of incidence the shift $S$ is

$S=(1-\frac{1}{\mu })\Delta t$

where $\mu$ is the refractive index, and $\Delta t$ is the thickness.

So the total shift due to all the slabs will be given as,

$S=S_1+S_2+S_3+...+S_n$

$S=\Delta t_1[1-\frac{1}{\mu }]+\Delta t_2[1-\frac{1}{\mu }]+...+\Delta t_n[1-\frac{1}{\mu }]$

As the thickness of the slabs is given as $1\text{cm},2\text{cm},...n\text{cm}$ we have

$S=[1-\frac{1}{\mu }](1+2+...+n)$

Now the refractive index is given as $\mu =1.5=\frac{3}{2}$ and the total shift is given as $S=5\text{cm}$

$5=(1-\frac{2}{3})\left(\frac{n(n+1)}{2}\right)$

$30=n(n+1)$

$5\times 6=n(n+1)$

$\Rightarrow n=5$