How far from a convex lens of focal length 20 cm would you place an object to get a virtual image, which is magnified 3 times ?

Options

A. 7.5 cm
B. 10.8 cm
C. 13.33 cm
D. 16.5 cm

Explanation

To solve the problem of how far from a convex lens of focal length 20 cm you would place an object to get a virtual image that is magnified 3 times, we can follow these steps:

### Step 1: Understand the Magnification Formula
The magnification (m) of a lens is given by the formula:
\[ m = \frac{h_i}{h_o} = \frac{v}{u} \]
where:
- \( h_i \) = height of the image
- \( h_o \) = height of the object
- \( v \) = image distance
- \( u \) = object distance

Given that the magnification is 3, we have:
\[ m = 3 \]
Thus, we can write:
\[ \frac{v}{u} = 3 \]
This means:
\[ v = 3u \]

### Step 2: Apply the Sign Convention
For a convex lens:
- The object distance (u) is taken as negative (since the object is placed on the same side as the incoming light).
- The image distance (v) will be positive for a virtual image.

Thus, we can express:
\[ u = -x \]
and since \( v = 3u \):
\[ v = 3(-x) = -3x \]

### Step 3: Use the Lens Formula
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Substituting the values we have:
- \( f = 20 \) cm (positive for a convex lens)
- \( v = -3x \)
- \( u = -x \)

So, substituting these into the lens formula:
\[ \frac{1}{20} = \frac{1}{-3x} - \frac{1}{-x} \]

### Step 4: Simplify the Equation
Rearranging the equation gives:
\[ \frac{1}{20} = -\frac{1}{3x} + \frac{1}{x} \]
Combining the fractions on the right:
\[ \frac{1}{20} = \frac{-1 + 3}{3x} = \frac{2}{3x} \]

### Step 5: Solve for x
Now, equate and solve for x:
\[ \frac{1}{20} = \frac{2}{3x} \]
Cross-multiplying gives:
\[ 3x = 40 \]
Thus:
\[ x = \frac{40}{3} \approx 13.33 \text{ cm} \]

### Conclusion
The object should be placed approximately **13.33 cm** from the convex lens to obtain a virtual image that is magnified 3 times.

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