Question

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

The equivalent refractive index of the system which will allow the image to be formed at the same place is $\mu=\frac{\sum_{i=1}^{k} t_{i}}{\sum_{i=1}^{k} \frac{t_{1}}{\mu_{1}}}$ .

Explanation:

Step 1 :

Given data  in the question  

Total number of slabs = k

The thickness is = $t_{1}, t_{2}, t_{3}, \ldots \ldots . t_{k}$

The Slabs refractive indices are = $\mu_{1}, \mu_{2}, \mu_{3}, \mu_{4} \dots \dots \dots \mu_{k}$

$\Delta t=\left(1-\frac{1}{\mu_{1}}\right) t_{1}+\left(1-\frac{1}{\mu_{2}}\right) t_{2}+\left(1-\frac{1}{\mu_{3}}\right) t_{3} \ldots \ldots \ldots \ldots+\left(1-\frac{1}{\mu_{k}}\right) t_{k}$ ---> eqn (1)

Step 2 :

μ the refractive index of the slabs and image combination is generated at the same spot,

$\Delta t=\left(1-\frac{1}{\mu}\right)\left(t_{1}+t_{2}+t_{3}+\cdots \ldots \ldots+t_{k}\right)$  ---> eqn (2)

Step 3 :

We get Equation (1) and (2)

$\left(1-\frac{1}{\mu}\right)\left(t_{1}+t_{2}+t_{3}+\cdots \ldots \ldots+t_{k}\right)=\left(1-\frac{1}{\mu_{1}}\right) t_{1}+\left(1-\frac{1}{\mu_{2}}\right) t_{2}+\left(1-\frac{1}{\mu_{3}}\right) t_{3} \ldots \ldots \ldots+\left(1-\frac{1}{\mu_{k}}\right) t_{k}$    $=\left(t_{1}+t_{2}+t_{3}+\cdots \ldots \ldots+t_{k}\right)-\left(\frac{t_{1}}{\mu_{1}}+\frac{t_{2}}{\mu_{2}}+\frac{t_{3}}{\mu_{3}}+\cdots \ldots \ldots \ldots+\frac{t_{k}}{\mu_{k}}\right)$

   $=-\frac{1}{\mu} \sum_{i=1}^{k} t_{1}$

   $=-\sum_{i=1}^{k} \frac{t_{1}}{\mu_{1}}$

$\mu=\frac{\sum_{i=1}^{k} t_{i}}{\sum_{i=1}^{k} \frac{t_{1}}{\mu_{1}}}$

Thus, the refractive index $\mu=\frac{\sum_{i=1}^{k} t_{i}}{\sum_{i=1}^{k} \frac{t_{1}}{\mu_{1}}}$ .