To determine the image distance produced by the first converging lens, we can use the lens formula:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
where:
- \( f \) is the focal length of the lens,
- \( v \) is the image distance,
- \( u \) is the object distance (taken as negative when using the sign convention).
Given:
- Focal length (\( f \)) = 30 cm (positive for a converging lens),
- Object distance (\( u \)) = -7 cm (since the distance is measured from the lens to the object).
Now, plugging these values into the lens formula:
\[ \frac{1}{30} = \frac{1}{v} - \frac{1}{-7} \]
This simplifies to:
\[ \frac{1}{30} = \frac{1}{v} + \frac{1}{7} \]
Rearranging gives:
\[ \frac{1}{v} = \frac{1}{30} - \frac{1}{7} \]
Finding a common denominator (which is 210):
\[ \frac{1}{30} = \frac{7}{210}, \quad \frac{1}{7} = \frac{30}{210} \]
Thus,
\[ \frac{1}{v} = \frac{7}{210} - \frac{30}{210} = \frac{-23}{210} \]
Taking the reciprocal to find \( v \):
\[ v = -\frac{210}{23} \approx -9.13 \text{ cm} \]
This negative sign indicates that the image is formed on the same side as the object (real image).
For the Second Lens
Now, if the object is moved 25 cm further away from the first lens, the new object distance becomes:
\[ u' = - (7 + 25) = -32 \text{ cm} \]
Assuming that the second converging lens is placed at the focal point of the first lens, the image formed by the first lens will now serve as the object for the second lens. The image distance of the first lens was approximately -9.13 cm. Thus, the new object distance for the second lens will be:
The distance from the first lens to the second lens is:
\[ d = 30 \text{ cm} - (-9.13) \approx 39.13 \text{ cm} \]
Now, placing the second lens at the position of the first lens (with a distance from the original object), the effective object distance for the second lens can be calculated as:
\[ u' = d - 39.13 \text{ cm} = -32 \text{ cm} \]
Now using the lens formula, we need to determine the focal length of the second lens. Let \( f' \) denote the focal length of the second lens, and we set our lens formula:
\[ \frac{1}{f'} = \frac{1}{v'} - \frac{1}{u'} \]
Assuming we want the image to be sharp, set \( v' \) equal to the distance of the image screen where the image should be formed. As we don't have that, but we know that the image should be formed at a distance equal to the new \( u' \).
If we want a sharp image, we change our equation and for clarity, we can assume:
For a converging lens, if the object is at \( u' \) and the image is formed at the same distance on the other side (for simplification as it should focus sharply):
Assuming that \( v' \) is also approximately around double the position for simplicity, we can say approach like:
\[ \frac{1}{f'} = \frac{1}{32} - \frac{1}{negative\ distance\ v'} \]
We can calculate and set these if we need. However generally, in similar scenarios if they indeed must meet, it's also plausible that the focal length required is 30 cm again in direct theory; determining again would also need experiment facts or gathering where significantly noted.
For precise finding for standard simplifications:
However, without precise image measures required, we can initially denote that characteristics on further lenses would require known measures or would set approximate 30 cm again as images touch through setups where converging would to meet sharp portrayals overall.