A thin symmetric lens provides an image of a fingerprint with a magnification of `+0.2` when the fingerprint is `1.0cm` farther from the lens than the focal point of the lens. What are the type and orientation of the image, and what is the type of lens ?

Answer

To solve the problem step by step, we will analyze the information provided and apply the principles of optics.

### Step 1: Understand the given information
We are given:
- Magnification (m) = +0.2
- The object (fingerprint) is located 1.0 cm farther from the lens than its focal point.

### Step 2: Determine the object distance (u)
Let the focal length of the lens be denoted as \( f \). Since the object is 1.0 cm farther than the focal point, we can express the object distance as:
\[ u = f + 1 \]
(Note: In the lens formula, the object distance is taken as negative, so we will use \( u = -(f + 1) \).)

### Step 3: Use the magnification formula
The magnification (m) for a lens is given by the formula:
\[ m = \frac{h'}{h} = -\frac{v}{u} \]
Where:
- \( h' \) = height of the image
- \( h \) = height of the object
- \( v \) = image distance
- \( u \) = object distance

Given that \( m = +0.2 \), we can write:
\[ 0.2 = -\frac{v}{u} \]

### Step 4: Substitute the expression for \( u \)
Substituting \( u = -(f + 1) \) into the magnification equation:
\[ 0.2 = -\frac{v}{-(f + 1)} \]
This simplifies to:
\[ 0.2 = \frac{v}{f + 1} \]
From this, we can express \( v \):
\[ v = 0.2(f + 1) \]

### Step 5: Determine the type and orientation of the image
Since the magnification is positive (\( +0.2 \)), this indicates that the image is erect and has the same orientation as the object. Additionally, a positive magnification suggests that the image is virtual.

### Step 6: Determine the type of lens
To determine the type of lens, we note that:
- The object is located between the focal point and the lens (since \( u \) is negative and \( |u| > |f| \)).
- A concave lens (diverging lens) produces a virtual image when the object is placed between the focal point and the lens.

### Conclusion
1. The image is virtual and erect.
2. The type of lens is concave (diverging lens).