Let the actual height of water filled be $h$ cm.

Since the student is looking from above, the water-filled part appears shallower due to refraction.

For a liquid of refractive index $\mu$, the apparent depth is $\text{apparent depth} = \frac{\text{real depth}}{\mu}$

Here, $\mu = \frac{4}{3}$ so the apparent depth of water is $\frac{h}{4/3} = \frac{3h}{4}$

The tumbler has total inner depth $17.5$ cm. The upper empty part has real depth $17.5 - h$ Since this part is air, its apparent depth remains the same: $17.5 - h$

The student feels the tumbler is half filled, i.e. the apparent water depth equals the apparent empty depth.

So, $\frac{3h}{4} = 17.5 - h$

Solve: $\frac{3h}{4} + h = 17.5$ $\frac{7h}{4} = 17.5$ $h = 17.5 \times \frac{4}{7} = 10$

Therefore, the actual height of water is $\boxed{10\text{ cm}}$

Option check:

• A: $11.7$ cm $\rightarrow$ incorrect
• B: $10$ cm $\rightarrow$ correct
• C: $7.5$ cm $\rightarrow$ incorrect
• D: $8.75$ cm $\rightarrow$ incorrect