https://file.aiquickdraw.com/mathbot/file/172703690292906ndi8bq.jpg

To solve this problem, we'll use the lens formula and magnification concept. Here’s a step-by-step solution:

### Step 1: Understanding the Problem
- **Focal Length (f)**: \(-2 \, \text{cm}\) (since it's a diverging lens)
- **Magnification (m)**: \(\frac{\text{image height}}{\text{object height}} = 0.4\) (given as 40%)
- **Characteristics of the Image**: Virtual and upright, which implies the magnification is positive.

### Step 2: Lens Formula
The lens formula is:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
where:
- \( f \) is the focal length of the lens.
- \( v \) is the image distance (negative for virtual images).
- \( u \) is the object distance (to be found, typically negative for real objects on the same side as the object).

### Step 3: Magnification Formula
Magnification (\(m\)) is given by:
\[ m = \frac{v}{u} \]
Given \( m = 0.4 \) (positive because the image is virtual and upright), we have:
\[ 0.4 = \frac{v}{u} \Rightarrow v = 0.4u \]

### Step 4: Substitute \( v \) in Lens Formula
Substituting \( v = 0.4u \) and \( f = -2 \, \text{cm} \) into the lens formula:
\[ \frac{1}{-2} = \frac{1}{0.4u} - \frac{1}{u} \]
\[ -\frac{1}{2} = \frac{1}{0.4u} - \frac{1}{u} \]

### Step 5: Simplify the Equation
Rewrite the equation to have a common denominator:
\[ -\frac{1}{2} = \frac{1 - 0.4}{0.4u} \]
\[ -\frac{1}{2} = \frac{0.6}{0.4u} \]
\[ -\frac{1}{2} = \frac{3}{2u} \]
Cross-multiplying to solve for \( u \):
\[ -u = 3 \]
\[ u = -3 \, \text{cm} \]

### Conclusion
The object distance \( u \) is \(-3 \, \text{cm}\). This negative sign indicates that the object is placed on the same side of the lens as the incoming light, which is typical for real objects.

Do you want details or have any questions?

Here are some related questions that might interest you:
1. What is the nature of the image formed by a converging lens when the object is placed at twice the focal length?
2. How does the focal length of a lens change if it is submerged in water?
3. What is the relationship between the radius of curvature of a lens and its focal length?
4. How do you determine the power of a lens and what does it signify?
5. How does spherical aberration affect the quality of the image produced by a lens?

**Tip:** When dealing with lens problems, always keep track of the sign conventions for distances and focal lengths to avoid errors in calculations.

A diverging lens of focal length -2 cm projects the image of an object onto a wall. What is the object distance if the image is virtual, upright, and 40% of the object size?

Learn how to calculate the object distance for a diverging lens with a focal length of -2 cm. The image is virtual, upright, and 40% of the object's size. This step-by-step solution covers key concepts in optics, including the lens formula and magnification.

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