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The problem asks for the distance \( D \) behind the mirror where the image of an object \( O \) is formed. Here's the key information provided:

- The object \( O \) is located 10 cm above the water level.
- The water column in the beaker is 12 cm.
- The mirror is below the water column.
- The refractive index of water \( \mu_w = \frac{4}{3} \).

### Step-by-Step Solution:

1. **Image of the object above water level**:
   The object is placed 10 cm above the water. This part is in air and doesn't undergo refraction. So, this part is directly reflected in the mirror. Hence, the distance of the image due to the air part is:
   \[
   D_{\text{air}} = 10 \, \text{cm}
   \]

2. **Image of the object below water level (water column)**:
   The image due to the portion below the water level undergoes refraction. Since the refractive index of water is given as \( \mu_w = \frac{4}{3} \), the apparent distance of the object when viewed from above (looking through the water) is given by the formula:
   \[
   d_{\text{apparent}} = \frac{d_{\text{real}}}{\mu_w}
   \]
   Where \( d_{\text{real}} \) is the real depth of the water, which is 12 cm.

   Substituting the values:
   \[
   d_{\text{apparent}} = \frac{12}{\frac{4}{3}} = 9 \, \text{cm}
   \]

3. **Total distance of the image behind the mirror**:
   The total distance of the image \( D \) behind the mirror will be the sum of the image distances due to both the air and water parts. This is calculated as:
   \[
   D = D_{\text{air}} + d_{\text{apparent}} = 10 \, \text{cm} + 9 \, \text{cm} = 19 \, \text{cm}
   \]

Thus, the distance \( D \) behind the mirror where the image is formed is **19 cm**.

### Answer:
The correct option is **(3) 19 cm**.

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Would you like further clarification or details? Here are some related questions:

1. How does the refractive index of a medium affect the apparent depth of an object?
2. Can you explain how mirror reflection works for objects placed in different media (like water and air)?
3. How would the image distance change if the object were placed entirely underwater?
4. What happens to the image if the refractive index of water were higher, say \( \mu_w = \frac{5}{3} \)?
5. How can you determine the real depth of an object if only its apparent depth and refractive index are known?

**Tip**: When dealing with refraction, always remember that light bends towards the normal when moving from a less dense to a denser medium, and away from the normal when moving from denser to less dense!

In the diagram shown, an object O is at a distance 10 cm above the water level. Water column in beaker is of 12 cm which is kept on a plane mirror. If refractive index of water is μw = 4/3, then the distance D behind the mirror where the image I is formed.

Solve a physics problem involving refraction and reflection. Find the distance D behind a mirror where an image is formed for an object 10 cm above water, with a water column of 12 cm, and a refractive index of 4/3. Suitable for Grades 10-12.

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