B

Given:
- The container has a total depth of H filled with two immiscible transparent liquids.
- Each liquid has a depth of H/2.
- The refractive indices are µ1 for the first liquid and µ2 for the second liquid.

Step 1: According to the formula for apparent depth when light passes through a medium of different refractive indices, we can derive the apparent depth for each liquid. The apparent depth can be given by the formula:
$$ d' = \frac{d}{\mu} $$
where d is the real depth and µ is the refractive index of the liquid.

Step 2: For the first liquid (depth = H/2, refractive index = µ1):
$$ d'_{1} = \frac{H/2}{µ1} $$

Step 3: For the second liquid (depth = H/2, refractive index = µ2):
$$ d'_{2} = \frac{H/2}{µ2} $$

Step 4: The total apparent depth from above is the sum of the apparent depths of both liquids:
$$ d'_{total} = d'_{1} + d'_{2} = \frac{H/2}{µ1} + \frac{H/2}{µ2} $$

Step 5: Factor out common terms:
$$ d'_{total} = \frac{H}{2} \left( \frac{1}{µ1} + \frac{1}{µ2} \right) $$

Step 6: Thus, the correct expression for the apparent depth of the vessel when viewed from above is:
$$ d'_{total} = \frac{H}{2} \left( \frac{1}{µ1} + \frac{1}{µ2} \right) $$

Conclusion: Therefore, the apparent depth from above is:
$$ \frac{H}{2} \left( \frac{1}{µ1} + \frac{1}{µ2} \right) $$
The answer matches with option B, confirming this as the correct solution.