A ray incident on the droplet of water at an angle of incidence i undergoes two reflections (not total) and emerges. If the deviation suffered by the ray within the drop is minimum and the refractive index of the droplet be `mu,` then show that `cos i =sqrt((u^2-1))/8.`

Explanation

To solve the problem, we will analyze the situation step-by-step and derive the required relationship for the angle of incidence \( i \) in terms of the refractive index \( \mu \).

### Step 1: Understanding the Geometry of Refraction and Reflection
When a ray of light enters a water droplet, it refracts at the surface, undergoes two reflections within the droplet, and then refracts again as it exits.

### Step 2: Define Angles
Let:
- \( i \) = angle of incidence
- \( r \) = angle of refraction

According to Snell's law:
\[ \sin i = \mu \sin r \]

### Step 3: Calculate Total Deviation
The total deviation \( D \) of the ray can be expressed as:
\[ D = \text{(deviation due to refraction)} + \text{(deviation due to reflection)} \]
The deviation due to refraction at the entry is \( i - r \) and at the exit is also \( i - r \). The deviation due to reflection is \( 180^\circ - 2r \) for each reflection. Thus, the total deviation becomes:
\[ D = 2(i - r) + 2(180^\circ - 2r) = 2(i - r) + 360^\circ - 4r \]
Simplifying this gives:
\[ D = 2i - 6r + 360^\circ \]

### Step 4: Minimize the Deviation
To find the angle \( i \) that minimizes the deviation \( D \), we differentiate \( D \) with respect to \( i \) and set the derivative to zero:
\[ \frac{dD}{di} = 2 - 6\frac{dr}{di} = 0 \]
From this, we find:
\[ \frac{dr}{di} = \frac{1}{3} \]

### Step 5: Applying Snell's Law
Using Snell's law at the point of incidence:
\[ \sin i = \mu \sin r \]
Differentiating both sides with respect to \( i \):
\[ \cos i = \mu \cos r \frac{dr}{di} \]
Substituting \( \frac{dr}{di} = \frac{1}{3} \):
\[ \cos i = \frac{\mu}{3} \cos r \]

### Step 6: Relating Angles Using Trigonometric Identities
Using the identity \( \sin^2 r + \cos^2 r = 1 \), we can express \( \cos r \) in terms of \( \sin r \):
\[ \cos r = \sqrt{1 - \sin^2 r} \]
From Snell's law, we have:
\[ \sin r = \frac{\sin i}{\mu} \]
Substituting this into the expression for \( \cos r \):
\[ \cos r = \sqrt{1 - \left(\frac{\sin i}{\mu}\right)^2} \]

### Step 7: Substitute and Rearrange
Substituting \( \cos r \) back into our equation:
\[ \cos i = \frac{\mu}{3} \sqrt{1 - \left(\frac{\sin i}{\mu}\right)^2} \]
Squaring both sides:
\[ \cos^2 i = \frac{\mu^2}{9} \left(1 - \frac{\sin^2 i}{\mu^2}\right) \]
This simplifies to:
\[ 9 \cos^2 i = \mu^2 - \sin^2 i \]
Using \( \sin^2 i = 1 - \cos^2 i \):
\[ 9 \cos^2 i + \cos^2 i = \mu^2 \]
Thus:
\[ 10 \cos^2 i = \mu^2 \]
Rearranging gives:
\[ \cos^2 i = \frac{\mu^2 - 1}{8} \]
Taking the square root:
\[ \cos i = \sqrt{\frac{\mu^2 - 1}{8}} \]

### Final Result
Thus, we have shown that:
\[ \cos i = \sqrt{\frac{\mu^2 - 1}{8}} \]