A person looking through a telescope focuses lens at a point on the edge of the bottom of an empty cylindrical vessel. Next he fills the entire vessel with a liquid oif refractive index `mu`, without disturbing the telescope. Now, he observes the mid point of the vessel. Determine the radius to depth ratio of the vessel

Options

A. `1/2sqrt((1-mu^(2))/(mu^(2)+1))`
B. `1/2sqrt((4-mu^(2))/(mu^(2)-1))`
C. `1/2sqrt((4+mu^(2))/(mu^(2)+1)`
D. `1/2sqrt((4+mu)/(mu+1))`

Answer

To solve the problem, we need to determine the radius to depth ratio of the cylindrical vessel after it has been filled with a liquid of refractive index \( \mu \). Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Geometry
- Consider a cylindrical vessel with depth \( h \) and radius \( r \).
- The person is looking through a telescope focused on the edge of the bottom of the vessel. When the vessel is filled with liquid, the light rays will bend due to refraction.

### Step 2: Identify the Angles
- Let \( i \) be the angle of incidence when light travels from air to the liquid.
- Let \( r \) be the angle of refraction in the liquid.
- According to Snell's Law, we have:
\[ n_1 \sin(i) = n_2 \sin(r) \]
where \( n_1 = 1 \) (refractive index of air) and \( n_2 = \mu \) (refractive index of the liquid).

### Step 3: Set Up the Geometry
- The light ray that is incident at the edge of the vessel will create a right triangle with the following sides:
- The opposite side (perpendicular) is \( 2r \) (the diameter of the vessel).
- The adjacent side (base) is \( \sqrt{(2r)^2 + h^2} \).

### Step 4: Apply Trigonometric Relationships
- From the geometry, we can express \( \sin(i) \) and \( \sin(r) \):
\[ \sin(i) = \frac{2r}{\sqrt{(2r)^2 + h^2}} \]
\[ \sin(r) = \frac{r}{\sqrt{r^2 + h^2}} \]

### Step 5: Apply Snell's Law
- Substituting into Snell's Law:
\[ 1 \cdot \frac{2r}{\sqrt{(2r)^2 + h^2}} = \mu \cdot \frac{r}{\sqrt{r^2 + h^2}} \]

### Step 6: Cross Multiply and Simplify
- Cross-multiplying gives:
\[ 2r \sqrt{r^2 + h^2} = \mu r \sqrt{(2r)^2 + h^2} \]
- Dividing both sides by \( r \) (assuming \( r \neq 0 \)):
\[ 2 \sqrt{r^2 + h^2} = \mu \sqrt{4r^2 + h^2} \]

### Step 7: Square Both Sides
- Squaring both sides results in:
\[ 4(r^2 + h^2) = \mu^2(4r^2 + h^2) \]
- Expanding gives:
\[ 4r^2 + 4h^2 = 4\mu^2 r^2 + \mu^2 h^2 \]

### Step 8: Rearranging the Equation
- Rearranging terms leads to:
\[ 4h^2 - \mu^2 h^2 = 4\mu^2 r^2 - 4r^2 \]
- Factoring out common terms:
\[ h^2(4 - \mu^2) = 4r^2(\mu^2 - 1) \]

### Step 9: Solve for the Radius to Depth Ratio
- Dividing both sides by \( h^2 \) gives:
\[ \frac{r^2}{h^2} = \frac{4 - \mu^2}{4(\mu^2 - 1)} \]
- Taking the square root:
\[ \frac{r}{h} = \frac{1}{2} \sqrt{\frac{4 - \mu^2}{\mu^2 - 1}} \]

### Final Result
- Therefore, the radius to depth ratio of the vessel is:
\[ \frac{r}{h} = \frac{1}{2} \sqrt{\frac{4 - \mu^2}{\mu^2 - 1}} \]