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Question: A straight rod is partially immersed in water \((\mu = 1.33)\). Its submerged position appears to be...

A straight rod is partially immersed in water (μ=1.33)(\mu = 1.33). Its submerged position appears to be inclined at 45{45^ \circ } with the surface when viewed vertically from air. What is the actual inclination of the rod?
A. 32.14{32.14^ \circ }
B. 45{45^ \circ }
C. 57.9{57.9^ \circ }
D. 62{62^ \circ }

Explanation

Solution

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
μ=sinisinr\mu = \dfrac{{\sin i}}{{\sin r}}
Where ii is angle of incidence and rr 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{90^ \circ }.

Complete step by step solution:
It is given that the refractive index of water is 1.33. That is,
μ=1.33\mu = 1.33
The angle of inclination of the submerged position as viewed vertically from air is given as 45{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
μ=sinisinr\mu = \dfrac{{\sin i}}{{\sin r}}
Where ii is angle of incidence and rr is angle of refraction.
Let us substitute the given values in the equation. Then, we get
1.33=sin45sinr1.33 = \dfrac{{\sin {{45}^ \circ }}}{{\sin r}}
1.33=12sinr\Rightarrow 1.33 = \dfrac{{\dfrac{1}{{\sqrt 2 }}}}{{\sin r}}
Since we know that, sin45=12\sin \,{45^ \circ } = \dfrac{1}{{\sqrt 2 }}
Thus,
sinr=121.33\sin r = \dfrac{{\dfrac{1}{{\sqrt 2 }}}}{{1.33}}
sinr=0.5316\therefore \sin \,r = 0.5316
We need the value of angle rr .Therefore, angle of refraction rr is
r=sin10.53=32.11r = {\sin ^{ - 1}}0.53 = {32.11^ \circ }
Actual inclination can be found by subtracting this angle from 90{90^ \circ }.
Therefore,
Actual inclination=9032=57.8957.9{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:
μ=sinisinr\mu = \dfrac{{\sin i}}{{\sin r}}
Where, μ\mu is the refractive index, ii is angle of incidence and rr is angle of refraction.