A converging lens of focal length 15 cm and a converging mirror of focal length 10 cm are placed 50 cm apart with common principal axis. A point source is placed in between the lens and the mirror at a distance of 40 cm from the lens. Find the locations of the two images formed.

Explanation

To solve the problem, we need to find the locations of the two images formed by a converging lens and a converging mirror placed 50 cm apart, with a point source located 40 cm from the lens.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Focal length of the lens (f₁) = +15 cm (positive for converging lens)
- Focal length of the mirror (f₂) = -10 cm (negative for concave mirror)
- Distance between the lens and the mirror = 50 cm
- Distance of the point source from the lens (object distance, u) = -40 cm (negative as per sign convention)

2. **Calculate the Image Formed by the Lens:**
- We use the lens formula:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
- Rearranging gives:
\[ \frac{1}{v} = \frac{1}{f} + \frac{1}{u} \]
- Substituting the values:
\[ \frac{1}{v} = \frac{1}{15} + \frac{1}{-40} \]
- Finding a common denominator (60):
\[ \frac{1}{v} = \frac{4}{60} - \frac{1.5}{60} = \frac{2.5}{60} \]
- Therefore:
\[ v = \frac{60}{2.5} = 24 \text{ cm} \]
- The first image is formed at 24 cm on the opposite side of the lens (positive value).

3. **Determine the Position of the First Image Relative to the Mirror:**
- Since the lens and mirror are 50 cm apart, the distance of the first image from the mirror is:
\[ \text{Distance from mirror} = 50 - 24 = 26 \text{ cm} \]
- The first image is located 26 cm in front of the mirror.

4. **Calculate the Object Distance for the Mirror:**
- The image formed by the lens acts as the object for the mirror. The object distance for the mirror (u') is:
\[ u' = -26 \text{ cm} \quad (\text{negative as it is in front of the mirror}) \]

5. **Calculate the Image Formed by the Mirror:**
- Using the mirror formula:
\[ \frac{1}{f} = \frac{1}{v'} - \frac{1}{u'} \]
- Rearranging gives:
\[ \frac{1}{v'} = \frac{1}{f} + \frac{1}{u'} \]
- Substituting the values:
\[ \frac{1}{v'} = \frac{1}{-10} + \frac{1}{-26} \]
- Finding a common denominator (130):
\[ \frac{1}{v'} = -\frac{13}{130} - \frac{5}{130} = -\frac{18}{130} \]
- Therefore:
\[ v' = -\frac{130}{18} \approx -7.22 \text{ cm} \]
- The negative sign indicates that the image is formed on the same side as the object, which is in front of the mirror.

### Summary of Results:
- The first image formed by the lens is at **24 cm** on the opposite side of the lens.
- The second image formed by the mirror is at approximately **7.22 cm** in front of the mirror.