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**Hint:**By taking the ratio of the velocity of light in glass and air medium, we can get the refractive index. By taking the ratio between the real depth and refractive index, we can get how much an ink dot appears to be raised.

**Formulas used:**

${\text{Refractive index = }}\dfrac{{{\text{Velocity of light in air}}}}{{{\text{Velocity of light in glass}}}}$

${\text{Refractive index = }}\dfrac{{{\text{Real depth}}}}{{{\text{Apparent depth}}}}$

**Complete step by step answer:**

A Refractive index can be found by taking the ratio velocity of light in glass and air.

$n = \dfrac{{{\text{Velocity of light in air}}}}{{{\text{Velocity of light in glass}}}}$

$\Rightarrow n = \dfrac{{3 \times {{10}^8}}}{{2 \times {{10}^8}}} = 1.5$

Here, we have found the refractive index $n = 1.5$

${\text{Refractive index = }}\dfrac{{{\text{Real depth}}}}{{{\text{Apparent depth}}}}$

$\Rightarrow n = \dfrac{6}{x} = 1.5$

$\therefore x = \dfrac{6}{{1.5}} = 4$

**So, the ink dot appears to be raised by $6 - 4 = 2{\text{cm}}$.**

**Additional information:**

Real Depth is the actual distance of an object beneath the surface, as would be measured by submerging a perfect ruler along with the object.Apparent depth in a medium is the depth of an object in a denser medium as seen from a rarer medium. Its value is smaller than real depth.

**Note:**The ratio between real and apparent depth is called a refractive index.The refractive index determines how much the path of light is bent when entering a material.This is described by Snell's law of refraction,

${n_1}\sin \sin {\theta _1} = {n_2}\sin \sin {\theta _2}$ ,

where ${\theta _1}$ and ${\theta _2}$ are the angles of incidence and refraction, respectively, of a ray crossing the interface between two media with refractive indices ${n_1}$ and ${n_2}$.

The refractive indices also determine the amount of light that is reflected when reaching the interface, as well as the critical angle for total internal reflection, their intensity, and Brewster's angle.

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