Question

k transparent slabs are arranged one over another. The refractive indices of the slabs are μ₁, μ₂, μ₃, ... μ_k and the thicknesses are t₁ t₂, t₃, ... 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

k number of transparent slabs are arranged one over the other.
Refractive indices of the slabs = μ₁, μ₂, μ₃, ..., μ_k
Thickness of the slabs = t₁, t₂, t₃,.., t_k
Shift due to one slab: \[∆ t = \left[ 1 - \frac{1}{\mu} \right]t\]
For the combination of multiple slabs, the shift is given by,
\[∆ t = \left[ 1 - \frac{1}{\mu_1} \right] t_1  + \left[ 1 - \frac{1}{\mu_2} \right] t_2  +  .  .  .  + \left[ 1 - \frac{1}{\mu_k} \right] t_k              .  .  . (i)\]|
Let μ be the refractive index of the combination of slabs.
The image is formed at the same place.
So, the shift will be:
\[\Delta t = \left[ 1 - \left( \frac{1}{\mu} \right) \right]( t_1  +  t_2    .  .  .  +  t_k )           .  .  . (ii)\] 
Equating (i) and (ii), we get:

\[\left[ 1 - \left( \frac{1}{\mu} \right) \right]( t_1  +  t_2    .  .  .  +  t_k )   = \left[ 1 - \frac{1}{\mu_1} \right] t_1  + \left[ 1 - \frac{1}{\mu_2} \right] t_2  + \left[ 1 - \frac{1}{\mu_k} \right] t_k \] 

\[ = ( t_1  +  t_2    .  .  .  +  t_k )   - \left( \frac{t_1}{\mu_1} + \frac{t_2}{\mu_2} + \frac{t_k}{\mu_k} \right)\] 

\[ =  - \frac{1}{\mu} \sum^k_{i = 1}  t_i  =  -  \sum^k_{i = 1} \left( \frac{t_i}{\mu_i} \right)\] 

\[ \Rightarrow \mu = \frac{\sum^k_{i = 1} t_i}{- \sum^k_{i = 1} \left( \frac{t_i}{\mu_i} \right)}\]

Hence, the required equivalent refractive index is
\[\frac{\sum^k_{i = 1} t_i}{\sum^k_{i = 1} \left( \frac{t_i}{\mu_i} \right)}\]