A pole light is situated near a still water pond. Calculate the angle subtended by the pole light to the pond surface such that light reflected from the water surface is completely polarised. Use `mu = 1.23`.

Answer

To solve the problem of calculating the angle subtended by the pole light to the pond surface such that light reflected from the water surface is completely polarized, we can follow these steps:

### Step 1: Understand the Condition for Complete Polarization
According to Brewster's Law, light is completely polarized upon reflection when the angle of incidence \( i \) satisfies the relationship:
\[ n = \tan(i) \]
where \( n \) is the refractive index of the medium (in this case, water).

### Step 2: Given Data
We are given that the refractive index \( n \) is 1.23.

### Step 3: Apply Brewster's Law
Using Brewster's Law:
\[ 1.23 = \tan(i) \]

### Step 4: Calculate the Angle of Incidence \( i \)
To find the angle \( i \), we take the arctangent (inverse tangent) of 1.23:
\[ i = \tan^{-1}(1.23) \]
Using a calculator, we find:
\[ i \approx 50.9^\circ \]

### Step 5: Calculate the Angle \( \theta \) with Respect to the Surface
The angle \( \theta \) that the pole light subtends with the pond surface is given by:
\[ \theta = 90^\circ - i \]
Substituting the value of \( i \):
\[ \theta = 90^\circ - 50.9^\circ = 39.1^\circ \]

### Final Answer
The angle subtended by the pole light to the pond surface such that light reflected from the water surface is completely polarized is:
\[ \theta \approx 39.1^\circ \]

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