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