A vessle of depth d is first filled to `( (2)/( 3))^(rd)` of its depth with a liquid of refractive index `( 3)/( 2)` and the rest of the vessel is filled with a liquid of refractive index `( 4)/( 3)`. What is the apparent depth of the inner surface of the bottom of the vessle.

Options

A. `(5d)/16`
B. `(29d)/36`
C. `(25d)/36`
D. `(15d)/16`

Explanation

To find the apparent depth of the inner surface of the bottom of the vessel filled with two different liquids, we can follow these steps:

### Step 1: Understand the Depths and Refractive Indices
The vessel has a total depth \( D \). It is filled to \( \frac{2}{3}D \) with a liquid of refractive index \( \mu_2 = \frac{3}{2} \) and the remaining \( \frac{1}{3}D \) with a liquid of refractive index \( \mu_1 = \frac{4}{3} \).

### Step 2: Calculate the Apparent Depth for the First Liquid
The apparent depth \( A_1 \) of the bottom of the vessel when viewed from the liquid of refractive index \( \mu_1 \) (the second liquid) can be calculated using the formula:
\[ A_1 = \frac{\text{Real Depth}}{\text{Refractive Index}} \]
The real depth for the first liquid is \( \frac{2}{3}D \) and the refractive index from the first liquid to the second liquid is given by:
\[ \mu_{21} = \frac{\mu_2}{\mu_1} = \frac{\frac{3}{2}}{\frac{4}{3}} = \frac{9}{8} \]
Thus, we can calculate \( A_1 \):
\[ A_1 = \frac{\frac{2}{3}D}{\frac{9}{8}} = \frac{2}{3}D \cdot \frac{8}{9} = \frac{16}{27}D \]

### Step 3: Calculate the Total Apparent Depth
Now, we need to find the total apparent depth \( A_2 \) when viewed from the top liquid (the second liquid). The real depth for this calculation is the depth of the second liquid plus the apparent depth \( A_1 \):
\[ \text{Real Depth} = \frac{1}{3}D + A_1 = \frac{1}{3}D + \frac{16}{27}D \]
To add these, we convert \( \frac{1}{3}D \) to a fraction with a denominator of 27:
\[ \frac{1}{3}D = \frac{9}{27}D \]
So,
\[ \text{Real Depth} = \frac{9}{27}D + \frac{16}{27}D = \frac{25}{27}D \]

Now, we apply the formula for apparent depth again:
\[ A_2 = \frac{\text{Real Depth}}{\text{Refractive Index of the second liquid}} \]
The refractive index of the second liquid (from the second liquid to air) is simply \( \mu_1 = \frac{4}{3} \):
\[ A_2 = \frac{\frac{25}{27}D}{\frac{4}{3}} = \frac{25}{27}D \cdot \frac{3}{4} = \frac{75}{108}D = \frac{25}{36}D \]

### Final Answer
The apparent depth of the inner surface of the bottom of the vessel is:
\[ \boxed{\frac{25}{36}D} \]