A paperweight ( μ = 1.5 ) shaped as a hemisphere of radius 3.0 cm...

For refraction at a spherical surface, the formula is:

$\dfrac{\mu_2}{v} - \dfrac{\mu_1}{u} = \dfrac{\mu_2 - \mu_1}{R}$

Here, light travels from the glass paperweight to the air. Therefore, the refractive index of the first medium is $\mu_1 = 1.5$ and the second medium is $\mu_2 = 1.0$.

The printed letters are located at the center of the flat surface, which is also the center of curvature of the spherical surface. Taking the pole of the spherical surface as the origin and the downward direction as negative, we have:

$u = -3.0 \text{ cm}$

$R = -3.0 \text{ cm}$

Substituting these values into the formula:

$\dfrac{1}{v} - \dfrac{1.5}{-3.0} = \dfrac{1.0 - 1.5}{-3.0}$

$\dfrac{1}{v} + 0.5 = \dfrac{-0.5}{-3.0}$

$\dfrac{1}{v} + \dfrac{1}{2} = \dfrac{1}{6}$

$\dfrac{1}{v} = \dfrac{1}{6} - \dfrac{1}{2} = -\dfrac{1}{3}$

$v = -3.0 \text{ cm}$

The image is formed at a distance of $3.0 \text{ cm}$ below the pole, which is exactly the original position of the printed page.

Conceptually, the light rays originating from the center of curvature strike the spherical surface normally and pass into the air undeviated. Thus, they appear to come from the object itself.

Since the image is formed at the object's exact location, the apparent shift is zero. Therefore, the letters appear at a height of $0 \text{ cm}$ above the page.