k transparent slabs are arranged one over another. The refractive indices of the slabs are g„ ...Ilk 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.

Explanation

To find the equivalent refractive index of a system of k transparent slabs arranged one over another, we can follow these steps:

### Step 1: Understand the Problem
We have k slabs, each with a thickness \( t_i \) and a refractive index \( \mu_i \). We need to find the equivalent refractive index \( \mu \) of the entire system such that the image formed through this combination is at the same place as it would be without the slabs.

### Step 2: Calculate the Shift for Each Slab
The shift \( S_i \) caused by each slab can be calculated using the formula:
\[ S_i = t_i \left(1 - \frac{1}{\mu_i}\right) \]
This formula accounts for the fact that light travels slower in a medium with a refractive index greater than 1.

### Step 3: Total Shift for the System
The total shift \( S \) for the entire system of k slabs is the sum of the individual shifts:
\[ S = S_1 + S_2 + S_3 + \ldots + S_k = \sum_{i=1}^{k} t_i \left(1 - \frac{1}{\mu_i}\right) \]

### Step 4: Equivalent Shift for a Single Slab
Now, if we replace the entire system of slabs with a single equivalent slab of thickness \( T \) (where \( T = t_1 + t_2 + \ldots + t_k \)) and refractive index \( \mu \), the shift for this equivalent slab can be expressed as:
\[ S = T \left(1 - \frac{1}{\mu}\right) \]

### Step 5: Set the Two Shifts Equal
To ensure that the image is formed at the same place, we set the total shift of the slabs equal to the shift of the equivalent slab:
\[ \sum_{i=1}^{k} t_i \left(1 - \frac{1}{\mu_i}\right) = T \left(1 - \frac{1}{\mu}\right) \]

### Step 6: Substitute \( T \)
Substituting \( T = t_1 + t_2 + \ldots + t_k \) into the equation gives:
\[ \sum_{i=1}^{k} t_i \left(1 - \frac{1}{\mu_i}\right) = (t_1 + t_2 + \ldots + t_k) \left(1 - \frac{1}{\mu}\right) \]

### Step 7: Rearranging the Equation
Rearranging the equation leads to:
\[ \sum_{i=1}^{k} t_i - \sum_{i=1}^{k} \frac{t_i}{\mu_i} = (t_1 + t_2 + \ldots + t_k) - \frac{(t_1 + t_2 + \ldots + t_k)}{\mu} \]

### Step 8: Solve for \( \mu \)
Now, we can solve for \( \mu \):
\[ \frac{(t_1 + t_2 + \ldots + t_k)}{\mu} = \sum_{i=1}^{k} \frac{t_i}{\mu_i} \]
Thus, we can express \( \mu \) as:
\[ \mu = \frac{(t_1 + t_2 + \ldots + t_k)}{\sum_{i=1}^{k} \frac{t_i}{\mu_i}} \]

### Final Answer
The equivalent refractive index \( \mu \) of the system is given by:
\[ \mu = \frac{T}{\sum_{i=1}^{k} \frac{t_i}{\mu_i}} \]
where \( T = t_1 + t_2 + \ldots + t_k \).

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