A printed page is pressed by a glass of water. The refractive index of the glass and water is 1.5 and 1.33, respectively. If the thickness of the bottom of glass is 1 cm and depth of water is 5 cm, how much the page will appear to be shifted if viewed from the top?

To solve this problem, we need to determine how much a printed page appears to shift when viewed from above through a glass of water. The glass has a refractive index of 1.5 and a bottom thickness of 1 cm, while the water has a refractive index of 1.33 and a depth of 5 cm. The page is pressed by the glass, meaning the arrangement from top to bottom is: air, water (5 cm deep), glass (1 cm thick), and then the page.

The apparent shift occurs due to refraction in both the water and the glass. The total shift is the difference between the real depth to the page and the apparent depth as seen from above. The real depth from the top to the page is the sum of the water depth and the glass thickness: $$5 \text{cm} + 1 \text{cm} = 6 \text{cm}.$$

The apparent depth is calculated by considering each medium separately. For the water layer, the apparent depth is the real depth divided by the refractive index of water: $$\frac{t_w}{\mu_w} = \frac{5}{1.33}.$$ For the glass layer, since it is viewed from within the water, we must use the relative refractive index of glass with respect to water. The relative refractive index is $$\frac{\mu_g}{\mu_w} = \frac{1.5}{1.33},$$ so the apparent depth for the glass is the real thickness multiplied by the ratio of the refractive indices: $$t_g \times \frac{\mu_w}{\mu_g} = 1 \times \frac{1.33}{1.5}.$$ Note that $$\frac{\mu_w}{\mu_g}$$ is used because the light is traveling from water to glass.

Therefore, the total apparent depth is the sum of these two apparent depths: $$\text{Apparent depth} = \frac{t_w}{\mu_w} + t_g \times \frac{\mu_w}{\mu_g} = \frac{5}{1.33} + 1 \times \frac{1.33}{1.5}.$$

Now, compute each term step by step. First, calculate $$\frac{5}{1.33}:$$ $$\frac{5}{1.33} = \frac{5 \times 100}{1.33 \times 100} = \frac{500}{133} \approx 3.7593984962406015.$$ Next, calculate $$\frac{1.33}{1.5}:$$ $$\frac{1.33}{1.5} = \frac{133}{150} = \frac{133 \div 1}{150 \div 1} = \frac{133}{150} \approx 0.8866666666666667.$$ Then, multiply by the glass thickness: $$1 \times 0.8866666666666667 = 0.8866666666666667.$$ Now, add the two apparent depths: $$3.7593984962406015 + 0.8866666666666667 = 4.646065162907268.$$

The apparent shift is the real depth minus the apparent depth: $$\text{Shift} = 6 - 4.646065162907268 = 1.353934837092732 \text{cm}.$$ Rounding this to four decimal places gives approximately 1.3539 cm, which matches option C (1.3533 cm) considering possible rounding differences in the calculation.

Hence, the correct answer is Option C.