A particle executes a simple harmonic motion of amplitude 1.0 cm along the principal axis of a convex lens of focal length 12 cm. The mean position of oscillation is at 20 cm from the lens. Find the amplitude of oscillation of the image of the particle.

Explanation

To solve the problem step by step, we will follow the outlined procedure:

### Step 1: Understand the Setup
We have a convex lens with a focal length (F) of 12 cm. A particle oscillates along the principal axis of the lens with an amplitude of 1.0 cm. The mean position of the oscillation is 20 cm from the lens.

### Step 2: Determine the Positions of the Particle
The mean position (M) is at 20 cm. The particle oscillates with an amplitude of 1.0 cm, so it moves from:
- Position A (to the right of M): 20 cm + 1 cm = 21 cm
- Position B (to the left of M): 20 cm - 1 cm = 19 cm

### Step 3: Calculate the Image Position for Position B
Using the lens formula:
\[ \frac{1}{v} = \frac{1}{f} + \frac{1}{u} \]
Where:
- \( v \) = image distance
- \( f \) = focal length = 12 cm
- \( u \) = object distance (negative as per sign convention)

For position B (u = -19 cm):
\[ \frac{1}{v_B} = \frac{1}{12} - \frac{1}{19} \]
Calculating the right-hand side:
\[ \frac{1}{v_B} = \frac{19 - 12}{12 \times 19} = \frac{7}{228} \]
Thus,
\[ v_B = \frac{228}{7} \approx 32.57 \text{ cm} \]

### Step 4: Calculate the Image Position for Position A
For position A (u = -21 cm):
\[ \frac{1}{v_A} = \frac{1}{12} - \frac{1}{21} \]
Calculating the right-hand side:
\[ \frac{1}{v_A} = \frac{21 - 12}{12 \times 21} = \frac{9}{252} = \frac{3}{84} \]
Thus,
\[ v_A = \frac{252}{9} = 28 \text{ cm} \]

### Step 5: Find the Amplitude of the Image Oscillation
The amplitude of the image oscillation can be calculated as:
\[ \text{Amplitude of image} = \frac{v_A - v_B}{2} \]
Substituting the values:
\[ \text{Amplitude of image} = \frac{28 - 32.57}{2} = \frac{-4.57}{2} = -2.285 \text{ cm} \]
Since amplitude is a positive quantity, we take the absolute value:
\[ \text{Amplitude of image} \approx 2.29 \text{ cm} \]

### Final Answer
The amplitude of oscillation of the image of the particle is approximately **2.29 cm**.

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