No, just changes the distance required between the lens and the screen in order to get a good picture. BUT it does not affect image quality.
The magnification of the telescope image is(focal length of the objective) divided by (focal length of the eyepiece).The focal length of the objective is fixed.Decreasing the focal length of the eyepiece increases the magnification of the image.(But it also makes the image dimmer.)
When the object lies within its focal length then no real image can be produced
Optical telescopes magnify an optical image - something that can be seen. They have lenses that magnify the image from light waves. they differ from Radio telescopes that detect waves other than the visual spectrum, these waves can not be "seen".
We don't think you can do it with that information. 'f-stop' = (focal length of the objective lens) divided by (its diameter) Magnification of the scope = (focal length of the objective) divided by (focal length of the eyepiece) Looks like in order to calculate the 'f-stop', you need to estimate or measure the focal length of either the objective or the eyepiece. Here's an idea: If you can temporarily separate the objective from the tube, use the objective to focus an image of the sun on the sidewalk. (Not on anything flammable.) Measure the distance from the lens to the sharpest image. With the 'object' at infinity, the image is at the focal length.
A positive lens is also called a magnifying lens. It has convex surfaces and it has a measureable focal length where it produces an inverted image of a distant object. The power in dioptres is the reciprocal of the focal length in metres.
The magnification of the telescope image is(focal length of the objective) divided by (focal length of the eyepiece).The focal length of the objective is fixed.Decreasing the focal length of the eyepiece increases the magnification of the image.(But it also makes the image dimmer.)
The magnification of the telescope image is(focal length of the objective) divided by (focal length of the eyepiece).The focal length of the objective is fixed.Decreasing the focal length of the eyepiece increases the magnification of the image.(But it also makes the image dimmer.)
1/object distance + 1/ image distance = 1/focal length
The focal length of the main optical system and the focal length of the lens forming the image.
Sum of reciprocal of object distance and reciprocal of image distance gives the reciprocal of focal length
Sum of reciprocal of object distance and reciprocal of image distance gives the reciprocal of focal length
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.
a camera? Or an eye? Something like that. Weird question
If an object's distance from the concave mirror is greater than the mirror's focal length, then the mirror image of it will be inverted. If the distance from the concave mirror is less than the focal length of the mirror, the image will not be inverted. No image will be produced if the distance from the mirror to the object is equal to the mirror's focal length.
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.
Focal Length
15cm