Spherical Lenses

To find the new size of the image when a concave lens is placed in the system, we need to consider how this additional lens affects the image formed by the convex lens alone.

Step 1: Image formation by the convex lens

The object is placed at infinity, and a convex lens of focal length is used. The image size formed by this convex lens is given as .

Thus, the convex lens forms an image at its focal length on the opposite side since the object is at infinity.

Step 2: Effect of the concave lens

The concave lens is placed from this image and has a focal length of (negative for concave lenses).

We will calculate the position of the final image using the lens formula:

Here, (the distance from the concave lens to the initial image), and .

Substituting these into the formula, we have:

Solving for gives:

The negative sign indicates that the image is formed on the same side as the object for the concave lens, which is virtual.

Step 3: Magnification and new image size

The magnification induced by the concave lens is given by:

The overall magnification of the system is the product of the magnifications by the convex lens and the concave lens:

(Since the initial image size was 2 cm, the magnification of the convex lens itself is effectively 1 here due to the object's position at infinity.)

The new image size would be:

2.5 \, \text{cm}.

Given Information:
Speed of object,
Speed of image,
Focal length of convex lens,

Step-by-Step Explanation:

Step 1: Relation between object speed and image speed

Using lens formula: where:
= object distance (negative for real objects),
= image distance (positive for real images).

Step 2: Differentiate lens formula with respect to time to find relation between velocities

Differentiating both sides:

Thus,

Given, , thus:

Thus, (since distances are positive for simplicity).

Step 3: Use in lens formula to find object distance

From lens formula:

But this leads to a contradiction if we directly substitute . Hence, let's carefully reconsider signs:

For a real object, is negative and is positive. Therefore, the correct condition considering signs becomes:

Substitute into lens formula carefully:

Thus:

Since , object distance is:

Thus, the object must be at from the lens.