A pin of length 2.00 cm is placed perpendicular to the principal axis of a converging lens. An inverted image of size 1.00 cm is formed at a distance of 40.0 cm from the pin. Find the focal length of the lens and its distance from the pin.

Explanation

To solve the problem step by step, we will use the lens formula and the magnification formula.

### Step 1: Understand the Problem
We have a pin of length 2.00 cm placed perpendicular to the principal axis of a converging lens. An inverted image of size 1.00 cm is formed at a distance of 40.0 cm from the pin. We need to find the focal length of the lens and its distance from the pin.

### Step 2: Define Variables
- Let the height of the object (pin) be \( h_o = 2.00 \, \text{cm} \).
- Let the height of the image be \( h_i = -1.00 \, \text{cm} \) (negative because the image is inverted).
- The distance between the pin and the image is given as \( 40.0 \, \text{cm} \).

### Step 3: Use the Magnification Formula
The magnification \( M \) is given by:
\[ M = \frac{h_i}{h_o} = \frac{-1}{2} \]
Also, magnification can be expressed in terms of image distance \( V \) and object distance \( U \):
\[ M = \frac{V}{U} \]
Setting these equal gives:
\[ \frac{V}{U} = \frac{-1}{2} \]
From this, we can express \( U \) in terms of \( V \):
\[ U = -2V \]

### Step 4: Set Up the Equation for Distances
According to the problem, the distance between the object (pin) and the image is 40 cm. Thus, we can write:
\[ -V + U = 40 \]
Substituting \( U = -2V \) into this equation:
\[ -V - 2V = 40 \]
This simplifies to:
\[ -3V = 40 \]
Thus,
\[ V = -\frac{40}{3} \, \text{cm} \]

### Step 5: Find Object Distance \( U \)
Now substituting \( V \) back to find \( U \):
\[ U = -2V = -2 \left(-\frac{40}{3}\right) = \frac{80}{3} \, \text{cm} \]

### Step 6: Use the Lens Formula
The lens formula is given by:
\[ \frac{1}{F} = \frac{1}{V} - \frac{1}{U} \]
Substituting the values of \( V \) and \( U \):
\[ \frac{1}{F} = \frac{1}{-\frac{40}{3}} - \frac{1}{\frac{80}{3}} \]
This simplifies to:
\[ \frac{1}{F} = -\frac{3}{40} - \frac{3}{80} \]
Finding a common denominator (80):
\[ \frac{1}{F} = -\frac{6}{80} - \frac{3}{80} = -\frac{9}{80} \]
Thus,
\[ F = -\frac{80}{9} \, \text{cm} \]
Since we are dealing with a converging lens, we take the positive value:
\[ F = \frac{80}{9} \, \text{cm} \]

### Step 7: Calculate the Distance of the Lens from the Pin
The distance of the lens from the pin can be calculated as:
\[ \text{Distance from pin} = |U| = \frac{80}{3} \, \text{cm} \]

### Final Answers
- Focal length of the lens \( F = \frac{80}{9} \, \text{cm} \)
- Distance of the lens from the pin \( = \frac{80}{3} \, \text{cm} \)