A fish is near the centre of a spherical water filled(`mu=4//3`)fish bowl, a child stands in air at a distance `2R` (`R` is the radius of curvature of the sphere)from the centre of the bowl. At what distance from the centre would the child nose appear to the fish situated at the centrer:

Options

A. R
B. 2R
C. 3R
D. 4R

Answer

To solve the problem, we need to find the distance from the center of a spherical fish bowl where a child's nose would appear to a fish located at the center of the bowl. The refractive index of the water in the bowl is given as \( \mu = \frac{4}{3} \). The child is standing at a distance of \( 2R \) from the center of the bowl, where \( R \) is the radius of curvature of the bowl.

### Step-by-Step Solution:

1. **Identify the Parameters**:
- The refractive index of water, \( \mu_2 = \frac{4}{3} \).
- The refractive index of air, \( \mu_1 = 1 \).
- The distance of the child from the center of the bowl, \( d = 2R \).

2. **Set Up the Refraction Formula**:
The formula for refraction at a spherical surface is given by:
\[ \frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R} \]
Where:
- \( v \) is the image distance (the distance from the center of the bowl where the child's nose appears to the fish).
- \( u \) is the object distance (the distance from the pole of the spherical surface to the object, which is the child).

3. **Determine the Object Distance \( u \)**:
The child is standing at a distance of \( 2R \) from the center of the bowl. The pole of the bowl is at the center, so:
\[ u = -R \]
(The negative sign indicates that the object is on the same side as the incident rays).

4. **Substituting Values into the Refraction Formula**:
Substitute \( \mu_1 \), \( \mu_2 \), \( u \), and \( R \) into the formula:
\[ \frac{\frac{4}{3}}{v} - \frac{1}{-R} = \frac{\frac{4}{3} - 1}{R} \]
Simplifying the right side:
\[ \frac{\frac{4}{3} - 1}{R} = \frac{\frac{4}{3} - \frac{3}{3}}{R} = \frac{\frac{1}{3}}{R} = \frac{1}{3R} \]

5. **Rearranging the Equation**:
The equation now becomes:
\[ \frac{\frac{4}{3}}{v} + \frac{1}{R} = \frac{1}{3R} \]
To eliminate the fractions, multiply through by \( 3Rv \):
\[ 4R + 3v = v \]
Rearranging gives:
\[ 4R = v - 3v \implies 4R = -2v \implies v = -2R \]

6. **Interpreting the Result**:
The negative sign indicates that the image (the child's nose) appears on the opposite side of the object (the child). Therefore, the distance from the center of the bowl to the image is \( 2R \) in the opposite direction.

7. **Calculating the Total Distance from the Center**:
Since the child is \( 2R \) away from the center and the image appears \( 2R \) in the opposite direction, the total distance from the center of the bowl is:
\[ R + 2R = 3R \]

### Final Answer:
The child's nose would appear to the fish at a distance of \( 3R \) from the center of the bowl.