A vessel of depth d is half filled with a liquid of refractive index `mu_1` and the other half is filled with a liquid of refractive index `mu_2`. The apparent depth of the vessel, when looked at normally, is

Options

A. `d(mu_1+mu_2)`
B. `d(1/mu_1+1/mu_2)`
C. `(d)/(2(mu_1+mu_2))`
D. `d/2(1/mu_1+1/mu_2)`

Explanation

To find the apparent depth of a vessel that is half-filled with two different liquids of refractive indices \( \mu_1 \) and \( \mu_2 \), we can follow these steps:

### Step 1: Understand the Setup
The vessel has a total depth \( d \). It is filled halfway with a liquid of refractive index \( \mu_1 \) and the other half with a liquid of refractive index \( \mu_2 \). Thus, the depth of each liquid is \( \frac{d}{2} \).

### Step 2: Calculate the Apparent Depth for Each Liquid
When light passes from a medium of one refractive index to another, it bends, causing objects to appear at a different depth than they actually are. The formula for apparent depth \( h' \) when viewed normally is given by:

\[ h' = \frac{h}{\mu} \]

where \( h \) is the actual depth and \( \mu \) is the refractive index of the liquid.

**For the first liquid (depth \( \frac{d}{2} \) and refractive index \( \mu_1 \)):**
\[ h_1' = \frac{\frac{d}{2}}{\mu_1} = \frac{d}{2\mu_1} \]

**For the second liquid (depth \( \frac{d}{2} \) and refractive index \( \mu_2 \)):**
\[ h_2' = \frac{\frac{d}{2}}{\mu_2} = \frac{d}{2\mu_2} \]

### Step 3: Combine the Apparent Depths
Since the two liquids are stacked, the total apparent depth \( H' \) of the vessel when viewed normally is the sum of the apparent depths of both liquids:

\[ H' = h_1' + h_2' = \frac{d}{2\mu_1} + \frac{d}{2\mu_2} \]

### Step 4: Factor Out Common Terms
We can factor out \( \frac{d}{2} \) from the equation:

\[ H' = \frac{d}{2} \left( \frac{1}{\mu_1} + \frac{1}{\mu_2} \right) \]

### Final Result
Thus, the apparent depth of the vessel when looked at normally is:

\[ H' = \frac{d}{2} \left( \frac{1}{\mu_1} + \frac{1}{\mu_2} \right) \]