A sphere made of transparent material of refractive index (`mu =3/2` )and of radius 50 cm has a small air bubble 10 cm below the surface. The apparent depth of the bubble if viewed from outside normally is

Options

A. `50/7` cm
B. `25/7 `cm
C. `100/7` cm
D. `75/7`cm

Answer

To find the apparent depth of the air bubble in the sphere made of transparent material, we can use the formula for refraction at a spherical surface. The formula we will use is derived from the lens maker's formula and is given by:

\[ \frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R} \]

Where:
- \( \mu_1 \) is the refractive index of the medium from which the light is coming (in this case, air, so \( \mu_1 = 1 \)).
- \( \mu_2 \) is the refractive index of the medium in which the object is located (the sphere, so \( \mu_2 = \frac{3}{2} \)).
- \( u \) is the object distance (the distance of the bubble from the surface of the sphere, which is \( -10 \) cm since it's below the surface).
- \( v \) is the image distance (the apparent depth we want to find).
- \( R \) is the radius of curvature of the sphere (which is \( -50 \) cm, as it is negative for a spherical surface).

### Step-by-step Solution:

1. **Identify the values:**
- \( \mu_1 = 1 \) (for air)
- \( \mu_2 = \frac{3}{2} \) (for the sphere)
- \( u = -10 \) cm (the bubble is below the surface)
- \( R = -50 \) cm (the radius of the sphere)

2. **Substitute the values into the formula:**
\[ \frac{\frac{3}{2}}{v} - \frac{1}{-10} = \frac{\frac{3}{2} - 1}{-50} \]

3. **Simplify the right side:**
\[ \frac{\frac{3}{2} - 1}{-50} = \frac{\frac{1}{2}}{-50} = -\frac{1}{100} \]

4. **Rewrite the equation:**
\[ \frac{\frac{3}{2}}{v} + \frac{1}{10} = -\frac{1}{100} \]

5. **Find a common denominator for the left side:**
\[ \frac{\frac{3}{2}}{v} + \frac{10}{100} = -\frac{1}{100} \]

6. **Combine the fractions:**
\[ \frac{\frac{3}{2}}{v} = -\frac{1}{100} - \frac{10}{100} = -\frac{11}{100} \]

7. **Cross-multiply to solve for \( v \):**
\[ \frac{3}{2} = -\frac{11v}{100} \]
\[ 3 \cdot 100 = -11v \cdot 2 \]
\[ 300 = -22v \]
\[ v = -\frac{300}{22} = -\frac{150}{11} \text{ cm} \]

8. **Calculate the apparent depth:**
The negative sign indicates that the image is virtual and located inside the sphere. The apparent depth is given by the absolute value:
\[ \text{Apparent Depth} = \frac{150}{11} \text{ cm} \approx 13.64 \text{ cm} \]

### Final Answer:
The apparent depth of the bubble when viewed from outside is approximately \( 13.64 \) cm.