A glass tumbler having inner depth of 17.5 cm is kept on a table. A student starts pouring water ($$\mu = \frac{4}{3}$$) into it while looking at the surface of water from the above. When he feels that the tumbler is half filled, he stops pouring water. Up to what height, the tumbler is actually filled?

We are told that the tumbler is $$17.5\ \text{cm}$$ deep. Suppose, after some pouring, the real (actual) height of the water column is $$h\ \text{cm}$$. Then the height of the empty air column that is still left above the water is obviously $$17.5-h\ \text{cm}$$.

The bottom of the tumbler is seen by the student through the layer of water. Whenever light travels from water into air, the bottom looks raised. The quantitative relation that gives this “lifting” is the well-known formula for normal viewing:

$$\text{apparent depth}=\frac{\text{real depth}}{\mu},$$

where $$\mu$$ is the refractive index of the liquid. Here the refractive index of water is given as $$\mu=\dfrac{4}{3}$$. Therefore the bottom appears to lie only

$$\frac{h}{\mu}=\frac{h}{\dfrac{4}{3}}=\frac{3h}{4}$$

centimetres below the water surface.

The student wants the tumbler to be half filled. Standing above the glass, he compares two vertical lengths that he can see at the same time:

• the empty air column of true height $$17.5-h\ \text{cm}$$, and
• the water column that appears to have a height $$\dfrac{h}{\mu}=\dfrac{3h}{4}\ \text{cm}$$.

He will stop pouring when, to his eye, these two lengths look equal, that is, when

$$\frac{h}{\mu}=17.5-h.$$

Substituting $$\mu=\dfrac{4}{3}$$ we obtain

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

so

$$\frac{3h}{4}=17.5-h.$$

Now multiply every term by $$4$$ to clear the denominator:

$$3h=70-4h.$$

Bring the $$h$$ terms together:

$$3h+4h=70,$$

$$7h=70.$$

Divide both sides by $$7$$:

$$h=\frac{70}{7}=10\ \text{cm}.$$

So the tumbler is actually filled with water up to a height of $$10\ \text{cm}$$, even though the student believes it to be half full.

Hence, the correct answer is Option A.