If 'f' is the focal length of the lens, and 'o' is the distance between the lens and the object, then the distance between the lens and the image is: ('f' times 'o') divided by ('o' minus 'f')
Easy way: Use it to form an image of the sun or moon, and measure the distance of the image behind the lens. When the object is at infinity, the distance between the lens and the image is the focal length of the lens.
concave lens formed virtual,erect and diminished image irrespective to the distance between the object and lens.
If the lens are thicker it affects the image distance.
If the lens are thicker it affects the image distance.
The image depends on the distance the object is from the lens.
Easy way: Use it to form an image of the sun or moon, and measure the distance of the image behind the lens. When the object is at infinity, the distance between the lens and the image is the focal length of the lens.
concave lens formed virtual,erect and diminished image irrespective to the distance between the object and lens.
If the lens are thicker it affects the image distance.
If the lens are thicker it affects the image distance.
The image depends on the distance the object is from the lens.
Find the smallest distance between the object and a real image, when the focal distance of the lens is2. Relevant equations, whereis the distance of the object from the lens andis the distance of the image.3. The attempt at a solutionI'm not even sure, what I'm trying to do here, since the definition of a real and virtual image is a bit vague to me. But I've got something...Letbe the desired distance. From the equation above we getso. Thenor. We get the same for, so the distance would be
It should appear the same.
The focal length of the lens and the distance between the lens and the object.
Concave lens (diverging) produces an upright image that is virtual. Although to create a real upright image would require 2 convex (converging) lens with a distance of their respective focal lengths between them.
1/(focal length) = 1/(distance of object) + 1/(distance of image) is the formula for calculating x of a lens knowing only the focal length which is the distance from the lens to the image of sun formed by it.
If the lens equation yields a negative image distance, then the image is a virtual image on the same side of the lens as the object. If it yields a negative focal length, then the lens is a diverging lens rather than the converging lens in the illustration.
Here's the equation you want. It's called the "Lensmaker's Formula".1/i + 1/o = 1/fi = image distance from the lenso = object distance from the lensf = focal length of the lens