Where should an object be placed from a converging lens of focal length 20 cm, so as to obtain a real image of magnification 2?

Options

A. 50 cm
B. 30 cm
C. `-50 cm`
D. `30 cm`

Explanation

To find the position of the object from a converging lens of focal length 20 cm that produces a real image with a magnification of 2, we can follow these steps:

### Step 1: Understand the Magnification
For a real image formed by a converging lens, the magnification (m) is given as negative. Since we want a magnification of 2, we write:
\[ m = -2 \]

### Step 2: Use the Magnification Formula
The magnification (m) for a lens is also defined as:
\[ m = \frac{v}{u} \]
where \( v \) is the image distance and \( u \) is the object distance.

### Step 3: Relate Image Distance to Object Distance
From the lens formula, we know:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Given that the focal length \( f \) is 20 cm, we can rearrange this to find \( v \):
\[ v = \frac{fu}{u + f} \]

### Step 4: Substitute Magnification into the Equation
Using the magnification formula:
\[ -2 = \frac{v}{u} \]
This implies:
\[ v = -2u \]

### Step 5: Substitute \( v \) in the Lens Formula
Now substitute \( v = -2u \) into the lens formula:
\[ \frac{1}{20} = \frac{1}{-2u} - \frac{1}{u} \]

### Step 6: Solve for \( u \)
To solve for \( u \), first find a common denominator:
\[ \frac{1}{20} = \frac{-1}{2u} + \frac{-1}{u} \]
This simplifies to:
\[ \frac{1}{20} = \frac{-1 - 2}{2u} \]
\[ \frac{1}{20} = \frac{-3}{2u} \]
Cross-multiplying gives:
\[ 2u = -60 \]
Thus:
\[ u = -30 \text{ cm} \]

### Conclusion
The object should be placed 30 cm in front of the lens (the negative sign indicates that the object is on the same side as the incoming light).

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