A point object is placed at a distance of 15 cm from a convex lens. The image is formed on the other side at a distance of 30 cm from the lens. When a concave lens is placed in contact with the convex lens, the image shifts away further by 30 cm. Calculate the focal lengths of the two lenses.

Explanation

To solve the problem, we need to find the focal lengths of the convex lens and the concave lens based on the given information.

### Step 1: Identify the given data for the convex lens
- Object distance (u) for the convex lens = -15 cm (the object is placed on the same side as the incoming light, hence negative)
- Image distance (v) for the convex lens = +30 cm (the image is formed on the opposite side of the lens, hence positive)

### Step 2: Use the lens formula for the convex lens
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Substituting the values:
\[ \frac{1}{f} = \frac{1}{30} - \frac{1}{-15} \]
\[ \frac{1}{f} = \frac{1}{30} + \frac{1}{15} \]
Finding a common denominator (which is 30):
\[ \frac{1}{f} = \frac{1}{30} + \frac{2}{30} = \frac{3}{30} = \frac{1}{10} \]
Thus, the focal length (f) of the convex lens is:
\[ f = 10 \text{ cm} \]

### Step 3: Analyze the effect of the concave lens
When the concave lens is placed in contact with the convex lens, the image shifts further away by 30 cm. Therefore, the new image distance (v') becomes:
\[ v' = 30 \text{ cm} + 30 \text{ cm} = 60 \text{ cm} \]

### Step 4: Use the lens formula for the combined system
Let the focal length of the concave lens be \( f_c \). The effective focal length \( f_{eff} \) of the combination of the two lenses is given by:
\[ \frac{1}{f_{eff}} = \frac{1}{f} + \frac{1}{f_c} \]
Since the new image distance (v') is 60 cm, we can use the lens formula again:
\[ \frac{1}{f_{eff}} = \frac{1}{60} - \frac{1}{15} \]
Finding a common denominator (which is 60):
\[ \frac{1}{f_{eff}} = \frac{1}{60} - \frac{4}{60} = -\frac{3}{60} = -\frac{1}{20} \]
Thus, the effective focal length \( f_{eff} \) is:
\[ f_{eff} = -20 \text{ cm} \]

### Step 5: Substitute known values to find the focal length of the concave lens
Now substituting \( f \) and \( f_{eff} \) into the equation:
\[ \frac{1}{-20} = \frac{1}{10} + \frac{1}{f_c} \]
Rearranging gives:
\[ \frac{1}{f_c} = -\frac{1}{20} - \frac{1}{10} \]
Finding a common denominator (which is 20):
\[ \frac{1}{f_c} = -\frac{1}{20} - \frac{2}{20} = -\frac{3}{20} \]
Thus, the focal length of the concave lens is:
\[ f_c = -\frac{20}{3} \text{ cm} \approx -6.67 \text{ cm} \]

### Final Results
- Focal length of the convex lens \( f = 10 \text{ cm} \)
- Focal length of the concave lens \( f_c \approx -6.67 \text{ cm} \)