A compound lens is made of two lenses having powers `+15.5 D` and `-5.5D`. An object of 3cm height is placed at a distance of 30 cm from this compound lens. Find the size of the image.

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Calculate the Power of the Compound Lens
The power of a lens is given in diopters (D). The total power \( P \) of a compound lens made of two lenses is the sum of their individual powers.

Given:
- Power of the first lens \( P_1 = +15.5 \, D \)
- Power of the second lens \( P_2 = -5.5 \, D \)

The total power \( P \) is calculated as:
\[ P = P_1 + P_2 = 15.5 - 5.5 = 10 \, D \]

### Step 2: Calculate the Focal Length of the Compound Lens
The focal length \( f \) of a lens in meters can be calculated from its power using the formula:
\[ f = \frac{100}{P} \, \text{(in cm)} \]
Substituting the total power:
\[ f = \frac{100}{10} = 10 \, \text{cm} \]

### Step 3: Determine the Object Distance
The object distance \( u \) is given as 30 cm. Since the object is placed in front of the lens, we take it as negative in the lens formula:
\[ u = -30 \, \text{cm} \]

### Step 4: Calculate the Magnification
The magnification \( m \) can be calculated using the formula:
\[ m = \frac{f}{f + u} \]
Substituting the values we have:
\[ m = \frac{10}{10 - 30} = \frac{10}{-20} = -\frac{1}{2} \]

### Step 5: Relate Magnification to Image Height
The magnification is also related to the heights of the object and the image:
\[ m = \frac{h_2}{h_1} \]
Where:
- \( h_1 \) is the height of the object (3 cm)
- \( h_2 \) is the height of the image

Rearranging the formula to find \( h_2 \):
\[ h_2 = m \cdot h_1 = -\frac{1}{2} \cdot 3 = -1.5 \, \text{cm} \]

### Conclusion
The size of the image is \( h_2 = -1.5 \, \text{cm} \). The negative sign indicates that the image is inverted.

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