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**Hint-**We know that according to Snell's law the refractive index of a medium is the ratio of sine of angle of incidence to the sine of angle of refraction. In equation form Snell's law is written as

$\mu = \dfrac{{\sin i}}{{\sin r}}$

Where $i$ is angle of incidence and $r$ is angle of refraction.

Using this we can find the angle of refraction.

Actual inclination can be found by subtracting this angle of refraction from ${90^ \circ }$.

**Complete step by step solution:**

It is given that the refractive index of water is 1.33. That is,

$\mu = 1.33$

The angle of inclination of the submerged position as viewed vertically from air is given as ${45^ \circ }$.

We need to find the actual inclination.

For this we should use Snell's law.

We know that according to Snell's law the refractive index of a medium is the ratio of sine of angle of incidence to the sine of angle of refraction. In equation form Snell's law is written as

$\mu = \dfrac{{\sin i}}{{\sin r}}$

Where $i$ is angle of incidence and $r$ is angle of refraction.

Let us substitute the given values in the equation. Then, we get

$1.33 = \dfrac{{\sin {{45}^ \circ }}}{{\sin r}}$

$ \Rightarrow 1.33 = \dfrac{{\dfrac{1}{{\sqrt 2 }}}}{{\sin r}}$

Since we know that, $\sin \,{45^ \circ } = \dfrac{1}{{\sqrt 2 }}$

Thus,

$\sin r = \dfrac{{\dfrac{1}{{\sqrt 2 }}}}{{1.33}}$

$\therefore \sin \,r = 0.5316$

We need the value of angle $r$ .Therefore, angle of refraction $r$ is

$r = {\sin ^{ - 1}}0.53 = {32.11^ \circ }$

Actual inclination can be found by subtracting this angle from ${90^ \circ }$.

Therefore,

Actual inclination=${90^ \circ } - {32^ \circ } = {57.89^ \circ } \cong {57.9^ \circ }$

**So, the correct answer is option C.**

**Note:**Using Snell's law what we get is angle of refraction. In order to find the actual inclination remember to subtract this angle from 90 degree.

Formula to remember:

$\mu = \dfrac{{\sin i}}{{\sin r}}$

Where, $\mu $ is the refractive index, $i$ is angle of incidence and $r$ is angle of refraction.

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