Question

$k$ transparent slabs are arranged one over another. The refractive indices of the slabs are ${\mu }_1,{\mu }_2,{\mu }_3,......{\mu }_k$ and the thicknesses are $t_1,t_2,t_3,..........t_k$. An object is seen through this combination with nearly perpendicular light. Find the equivalent refractive index of the system which will allow the image to be formed at the same place.

Answer 1

Total no. of slabs = k, thickness $=t_1,t_2,t_3....t_k$

Refractive index $={\mu }_1,{\mu }_2,{\mu }_3,{\mu }_4,...{\mu }_k$

$\therefore$ The shift $\Delta t=(1-\frac{1}{{\mu }_1})t_1+(1-\frac{1}{{\mu }_2})t_2+.....+(1-\frac{1}{{\mu }_k})t_k$ ___(1)

If, $\mu \to$ refractive index of combination of slabs and image is formed at same place,

$\Delta t=(1-\frac{1}{\mu })(t_1+t_2+...+t_k)$ __(2)

Equation(1) and (2), we get

$(1-\frac{1}{\mu })(t_1+t_2+...+t_k)=(1-\frac{1}{{\mu }_1})t_1+(1-\frac{1}{{\mu }_2})t_2+...+(1-\frac{1}{{\mu }_k})t_k$

$=(t_1+t_2+....+t_k)-(\frac{t_1}{{\mu }_1}+\frac{t_2}{{\mu }_2}+...+\frac{t_k}{{\mu }_k})$

$=-\frac{1}{\mu }\sum _{i=1}^k{t}_1=-\sum _{i=1}^k\left(\frac{t_1}{{\mu }_1}\right)\Rightarrow \mu =\frac{\sum _{i=1}^k{t}_i}{\sum _{i=1}^k\left(t_1/{\mu }_1\right)}$