A 5mm high pin is placed at a distance of 15cm from a convex lens of focal length 10cm. A second lens of focal length 5cm is placed 40cm from the first lens and 55cm from the pin. Find (a) the position of the final image, (b) its nature and (c) its size.

Sure, let's solve this step by step.

(a) First, we need to find the position of the image formed by the first lens. We can use the lens formula, which is 1/f = 1/v - 1/u. Here, f is the focal length of the lens, v is the image distance, and u is the object distance.

For the first lens, f1 = 10 cm and u1 = -15 cm (we take it as negative because the object is on the same side of the light source). Substituting these values in the lens formula, we get:

1/v1 = 1/f1 + 1/u1 1/v1 = 1/10 + 1/(-15) 1/v1 = (15 - 10) / (10 * 15) 1/v1 = 1/30 So, v1 = 30 cm.

This means the image formed by the first lens is 30 cm on the other side of the lens.

Now, the distance between the two lenses is 40 cm. So, this image will act as the object for the second lens and its distance from the second lens will be 40 - 30 = 10 cm.

Now, let's find the position of the image formed by the second lens. The focal length of the second lens, f2 = 5 cm and the object distance, u2 = -10 cm (again, negative because the object is on the same side of the light source).

Substituting these values in the lens formula, we get:

1/v2 = 1/f2 + 1/u2 1/v2 = 1/5 + 1/(-10) 1/v2 = (10 - 5) / (5 * 10) 1/v2 = 1/10 So, v2 = 10 cm.

This means the final image is formed 10 cm on the other side of the second lens. But since the second lens is 40 cm away from the first lens, the final image is at a distance of 40 + 10 = 50 cm from the first lens.

(b) The final image is real and inverted because it is formed on the opposite side of the light source.

(c) Now, let's find the size of the final image. The magnification m of a lens is given by the ratio of the image height to the object height, and it is also equal to -v/u (the negative sign indicates that the image is inverted).

First, let's find the magnification of the first lens:

m1 = -v1/u1 = -30/(-15) = 2.

This means the image formed by the first lens is twice the size of the object, so its height is 2 * 5 mm = 10 mm.

Now, this image acts as the object for the second lens. So, let's find the magnification of the second lens:

m2 = -v2/u2 = -10/(-10) = 1.

This means the final image is the same size as the object for the second lens, which is 10 mm. So, the final image is 10 mm high.