A container contains water up to a height of 20 cm and there is a point source at the centre of the bottom of the container. A rubber ring of radius r floats centrally on the water. The ceiling of the room is 2.0 m above the water surface. (a) Find the radius of the shadow of the ring formed on the ceiling if r = 15 cm. (b) Find the maximum value of r for which the shadow of the ring is formed on the ceiling. Refractive index of water =4/3.

Explanation

To solve the problem, we will break it down into two parts as specified in the question.

### Part (a): Find the radius of the shadow of the ring formed on the ceiling if r = 15 cm.

1. **Identify Given Values:**
- Height of water (h) = 20 cm = 0.20 m
- Radius of the rubber ring (r) = 15 cm = 0.15 m
- Height of the ceiling from the water surface = 2.0 m

2. **Calculate the Total Height from the Source to the Ceiling:**
- Total height (H) = Height of water + Height of ceiling = 0.20 m + 2.0 m = 2.20 m

3. **Use Snell's Law to Find the Angles:**
- The refractive index of water (n1) = 4/3
- The refractive index of air (n2) = 1
- According to Snell's Law:
\[ n_1 \sin(i) = n_2 \sin(r) \]
- This simplifies to:
\[ \frac{4}{3} \sin(i) = \sin(r) \]

4. **Calculate the Hypotenuse:**
- The distance from the point source to the edge of the ring is given by the Pythagorean theorem:
\[ \text{Hypotenuse} = \sqrt{(20 \text{ cm})^2 + (15 \text{ cm})^2} = \sqrt{400 + 225} = \sqrt{625} = 25 \text{ cm} \]

5. **Find \(\sin(i)\) and \(\sin(r)\):**
- The vertical distance (opposite side) to the point source is 15 cm (radius of the ring).
- The hypotenuse is 25 cm.
- Therefore:
\[ \sin(i) = \frac{15}{25} = \frac{3}{5} \]
- Substitute \(\sin(i)\) into Snell's Law:
\[ \frac{4}{3} \cdot \frac{3}{5} = \sin(r) \Rightarrow \sin(r) = \frac{4}{5} \]

6. **Calculate the Angle \(r\):**
- Use the relationship between sine and tangent:
\[ \tan(r) = \frac{\sin(r)}{\sqrt{1 - \sin^2(r)}} = \frac{\frac{4}{5}}{\sqrt{1 - \left(\frac{4}{5}\right)^2}} = \frac{\frac{4}{5}}{\sqrt{\frac{9}{25}}} = \frac{4}{5} \cdot \frac{5}{3} = \frac{4}{3} \]

7. **Calculate the Radius of the Shadow:**
- Let \(x\) be the radius of the shadow on the ceiling.
- Using the tangent of angle \(r\):
\[ \tan(r) = \frac{x/2}{2.20} \Rightarrow \frac{4}{3} = \frac{x/2}{2.20} \]
- Rearranging gives:
\[ x = \frac{4}{3} \cdot 2.20 \cdot 2 = \frac{4 \cdot 2.20}{3} = \frac{8.8}{3} \approx 2.93 \text{ m} \]

### Part (b): Find the maximum value of r for which the shadow of the ring is formed on the ceiling.

1. **Critical Angle Condition:**
- The critical angle \(C\) occurs when \(\sin(C) = 1\). This means all light is refracted at the maximum angle.
- Using Snell's Law at the critical angle:
\[ \sin(C) = \frac{r}{\sqrt{(20 \text{ cm})^2 + r^2}} = \frac{3}{4} \]

2. **Setting Up the Equation:**
- Rearranging gives:
\[ r = \frac{3}{4} \sqrt{(20)^2 + r^2} \]

3. **Squaring Both Sides:**
- Squaring gives:
\[ r^2 = \frac{9}{16} (400 + r^2) \]
- Rearranging leads to:
\[ 16r^2 = 9(400 + r^2) \Rightarrow 16r^2 = 3600 + 9r^2 \Rightarrow 7r^2 = 3600 \]

4. **Solving for r:**
- Thus:
\[ r^2 = \frac{3600}{7} \Rightarrow r = \sqrt{\frac{3600}{7}} \approx 2.67 \text{ cm} \]

### Final Answers:
- (a) The radius of the shadow of the ring formed on the ceiling is approximately **2.93 m**.
- (b) The maximum value of \(r\) for which the shadow of the ring is formed on the ceiling is approximately **2.67 cm**.