if an object of 5 cm height is placed at a distance of 15 cm from a convex lens of focal length 10 cm find the position nature and height of

The question seems to be incomplete. However, I assume you're asking for the position, nature, and height of the image formed by the convex lens.

We can use the lens formula to solve this:

1/v - 1/u = 1/f

Where: v = image distance u = object distance f = focal length

Given: u = -15 cm (we take it as negative because the object is on the same side of the light source) f = 10 cm

Substituting these values into the lens formula:

1/v - (-1/15) = 1/10 1/v = 1/10 + 1/15 1/v = (3 + 2) / 30 1/v = 5/30 1/v = 1/6

So, v = 6 cm. The positive sign indicates that the image is formed on the opposite side of the light source.

Next, we can find the height of the image using the magnification formula:

magnification, m = -v/u = image height/object height

Substituting the given values:

m = -6/-15 = image height/5 image height = 5 * (6/15) = 2 cm

The positive sign indicates that the image is erect.

So, the image is formed 6 cm from the lens, on the opposite side of the light source. The image is erect and 2 cm in height.