A vessel of depth 2d cm is half filled with a liquid of refractive index `mu_(1)` and the upper half with a liquid of refractive index `mu_(2)`. The apparent depth of the vessel seen perpendicularly is

Options

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

Answer

To solve the problem of finding 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 of \( 2d \) cm. The lower half (depth \( d \)) is filled with liquid of refractive index \( \mu_1 \), and the upper half (also depth \( d \)) is filled with liquid of refractive index \( \mu_2 \).

### Step 2: Identify the Apparent Depth
When viewed perpendicularly, the apparent depth of the vessel can be calculated using the concept of refraction. The apparent depth \( h' \) can be found by considering the refraction at the interface between the two liquids and at the surface of the liquid.

### Step 3: Apply Snell's Law
Using Snell's law at the interface between the two liquids:
\[ \mu_1 \sin I = \mu_2 \sin R \]
For small angles, we can approximate \( \sin I \approx I \) and \( \sin R \approx R \).

### Step 4: Relate Angles to Depths
Let \( x \) be the distance from the bottom of the vessel to the point where the ray of light exits the liquid of refractive index \( \mu_1 \). The geometry gives us:
- For the angle of incidence \( I \) at the interface:
\[ \tan I = \frac{x}{d} \]
- For the angle of refraction \( R \) in the liquid of refractive index \( \mu_2 \):
\[ \tan R = \frac{x}{d} \]

### Step 5: Substitute into Snell's Law
Substituting the values from the tangent relationships into Snell's law gives:
\[ \mu_1 \frac{x}{d} = \mu_2 \frac{x}{h} \]
Where \( h \) is the depth in the liquid of refractive index \( \mu_2 \).

### Step 6: Solve for \( h \)
Rearranging gives:
\[ h = \frac{\mu_1 d}{\mu_2} \]

### Step 7: Calculate the Apparent Depth
Now, since the observer is in air (refractive index \( \mu = 1 \)), we apply Snell's law again at the air-liquid interface:
\[ \frac{h'}{h} = \frac{1}{\mu_1} \]
Thus:
\[ h' = \frac{h}{\mu_1} = \frac{\mu_1 d}{\mu_2 \mu_1} = \frac{d}{\mu_2} + d \]

### Step 8: Final Expression for Apparent Depth
Combining these results, we find the total apparent depth \( h' \):
\[ h' = \frac{d}{\mu_2} + d = d \left(1 + \frac{1}{\mu_2}\right) \]

### Conclusion
The final expression for the apparent depth of the vessel seen perpendicularly is:
\[ h' = \frac{d}{\mu_2} + d \]