A convex lens produces a double size real image when an object is placed at a distance of 18 cm from it. Where should the object be placed to produce a triple size real image ?

Explanation

To solve the problem, we need to determine the position of the object that will produce a triple size real image using a convex lens. We will follow these steps:

### Step 1: Understand the Given Information
We know that:
- The object is initially placed at a distance of \( U = -18 \, \text{cm} \) (the negative sign indicates the object is on the same side as the incoming light).
- The magnification \( M \) for the first case is 2 (double size image).

### Step 2: Use the Magnification Formula
The magnification \( M \) is given by the formula:
\[ M = \frac{V}{U} \]
Where \( V \) is the image distance. For the first case:
\[ 2 = \frac{V}{-18} \]
From this, we can find \( V \):
\[ V = -2 \times 18 = -36 \, \text{cm} \]

### Step 3: Apply the Lens Formula
The lens formula is given by:
\[ \frac{1}{F} = \frac{1}{V} - \frac{1}{U} \]
Substituting the values we have:
\[ \frac{1}{F} = \frac{1}{-36} - \frac{1}{-18} \]
Finding a common denominator (which is 36):
\[ \frac{1}{F} = -\frac{1}{36} + \frac{2}{36} = \frac{1}{36} \]
Thus, the focal length \( F \) is:
\[ F = 36 \, \text{cm} \]

### Step 4: Determine the New Magnification for Triple Size Image
For the triple size image, the magnification \( M \) is 3. Using the magnification formula again:
\[ 3 = \frac{V_1}{U_1} \]
Where \( U_1 \) is the new object distance. Therefore:
\[ V_1 = 3U_1 \]

### Step 5: Use the Lens Formula Again
Substituting \( V_1 \) into the lens formula:
\[ \frac{1}{F} = \frac{1}{V_1} - \frac{1}{U_1} \]
Substituting \( F = 36 \, \text{cm} \) and \( V_1 = 3U_1 \):
\[ \frac{1}{36} = \frac{1}{3U_1} - \frac{1}{U_1} \]
Finding a common denominator for the right side:
\[ \frac{1}{36} = \frac{1 - 3}{3U_1} = \frac{-2}{3U_1} \]
Cross-multiplying gives:
\[ -2 \cdot 36 = 3U_1 \implies -72 = 3U_1 \implies U_1 = -24 \, \text{cm} \]

### Step 6: Conclusion
The object should be placed at a distance of \( 24 \, \text{cm} \) from the lens on the same side as the incoming light (indicated by the negative sign).

### Final Answer
The object should be placed at \( 24 \, \text{cm} \) from the convex lens.

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