A converging lens of focal length 15 cm and a converging mirror of focal length 10 cm are placed 50 cm apart. If a pin of length 2.0 cm is placed 30 cm from the lens farther away from the mirror, where will the final image form and what will be the size of the final image ?

Explanation

To solve the problem step by step, we will follow the sequence of events that occur when light passes through the converging lens and then reflects off the converging mirror.

### Step 1: Identify the given data
- Focal length of the converging lens (f1) = +15 cm (positive for a convex lens)
- Focal length of the converging mirror (f2) = -10 cm (negative for a concave mirror)
- Distance between the lens and the mirror = 50 cm
- Object distance from the lens (u) = -30 cm (since the object is placed on the same side as the incoming light)
- Height of the object (h) = 2.0 cm

### Step 2: Calculate the image distance from the lens using the lens formula
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Rearranging gives:
\[ \frac{1}{v} = \frac{1}{f} + \frac{1}{u} \]
Substituting the values:
\[ \frac{1}{v} = \frac{1}{15} + \frac{1}{-30} \]
Calculating the right side:
\[ \frac{1}{v} = \frac{2}{30} - \frac{1}{30} = \frac{1}{30} \]
Thus,
\[ v = 30 \text{ cm} \]

### Step 3: Determine the nature of the image formed by the lens
The positive value of \(v\) indicates that the image is formed on the opposite side of the lens and is real.

### Step 4: Calculate the magnification produced by the lens
The magnification (M) is given by:
\[ M = \frac{h'}{h} = \frac{v}{u} \]
Substituting the values:
\[ M = \frac{30}{-30} = -1 \]
This means the image is inverted and has the same size as the object. Therefore, the height of the image (h') is:
\[ h' = -1 \times 2 \text{ cm} = -2 \text{ cm} \]

### Step 5: Determine the object distance for the mirror
The distance between the lens and the mirror is 50 cm. Since the image formed by the lens is 30 cm from the lens, the distance from the image to the mirror is:
\[ \text{Distance from image to mirror} = 50 - 30 = 20 \text{ cm} \]
Thus, the object distance for the mirror (u') is:
\[ u' = -20 \text{ cm} \quad (\text{negative since it is on the same side as the incoming light}) \]

### Step 6: Calculate the image distance from the mirror using the mirror formula
The mirror formula is given by:
\[ \frac{1}{f} = \frac{1}{v'} - \frac{1}{u'} \]
Rearranging gives:
\[ \frac{1}{v'} = \frac{1}{f} + \frac{1}{u'} \]
Substituting the values:
\[ \frac{1}{v'} = \frac{1}{-10} + \frac{1}{-20} \]
Calculating the right side:
\[ \frac{1}{v'} = -\frac{1}{10} - \frac{1}{20} = -\frac{2}{20} - \frac{1}{20} = -\frac{3}{20} \]
Thus,
\[ v' = -\frac{20}{3} \approx -6.67 \text{ cm} \]

### Step 7: Determine the nature of the image formed by the mirror
The negative value of \(v'\) indicates that the image is virtual and located on the same side as the object for the mirror.

### Step 8: Calculate the magnification produced by the mirror
Using the magnification formula for the mirror:
\[ M' = -\frac{v'}{u'} \]
Substituting the values:
\[ M' = -\left(-\frac{20/3}{-20}\right) = \frac{1}{3} \]
Thus, the height of the final image (h'') is:
\[ h'' = M' \times h = \frac{1}{3} \times 2 \text{ cm} = \frac{2}{3} \text{ cm} \approx 0.67 \text{ cm} \]

### Final Answer:
The final image is formed approximately 6.67 cm in front of the mirror (on the same side as the object) and is virtual, inverted, and has a height of approximately 0.67 cm.