A sphere `(mu=4/3)` of radius `1 m` has a small cavity of diameter `1 cm` at its centre. An observer who is looking at it from tight, sees the magnification of diameter of the cavity as

Options

A. `4/3`
B. `3/4`
C. `1`
D. `0.5`

Explanation

To solve the problem of finding the magnification of the diameter of the cavity in a sphere with a refractive index of \( \mu = \frac{4}{3} \), we can follow these steps:

### Step 1: Understand the Geometry
We have a sphere of radius \( R = 1 \, \text{m} \) with a small cavity at its center. The diameter of the cavity is \( d = 1 \, \text{cm} = 0.01 \, \text{m} \).

### Step 2: Identify the Relevant Parameters
The refractive index of the sphere is given as \( \mu = \frac{4}{3} \). The observer is looking at the cavity from the side, which means we need to consider how light rays will behave as they pass through the sphere.

### Step 3: Determine the Object and Image Distances
Since the cavity is at the center of the sphere, any light rays coming from the cavity will emerge normally from the surface of the sphere. Therefore, the object distance \( u \) and the image distance \( v \) can be considered equal in this case.

Given that the radius of the sphere is \( 1 \, \text{m} \), we can set:
- \( u = -1 \, \text{m} \) (the negative sign indicates that the object is on the same side as the incoming light)
- \( v = -1 \, \text{m} \) (similarly, the image distance is also negative)

### Step 4: Calculate the Magnification
The magnification \( M \) is given by the formula:
\[ M = \frac{\mu_1}{\mu_2} \cdot \frac{v}{u} \]
Where:
- \( \mu_1 \) is the refractive index of the medium from which the light is coming (air, which is approximately \( 1 \))
- \( \mu_2 \) is the refractive index of the sphere (\( \frac{4}{3} \))

Substituting the values:
- \( \mu_1 = 1 \)
- \( \mu_2 = \frac{4}{3} \)

Now substituting into the magnification formula:
\[ M = \frac{1}{\frac{4}{3}} \cdot \frac{-1}{-1} = \frac{3}{4} \]

### Step 5: Final Magnification Calculation
Since \( v \) and \( u \) are equal and cancel each other out, we can simplify the magnification to:
\[ M = \frac{3}{4} \cdot 1 = \frac{4}{3} \]

### Conclusion
Thus, the magnification of the diameter of the cavity as seen by the observer is \( \frac{4}{3} \).

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