13.7 millimeters
13.7 millimeters
5.0 millimeters
A diverging lens can only produce a virtual image, because the light passing through a diverging lens never converges to a point. The virtual image produced by a diverging lens is always right-side-up and smaller than the original object. The image and the object viewed are always on the same side of the lens. Diverging lenses are used as viewfinders in cameras.
The distance between the object and mirror is 15 mm. The distance between the image and mirror is 15 mm. Therefore, the distance between the image and object is 15 mm plus 15 mm which equals 30 mm.
If a single lens forms a virtual image of an object, thenthe lens could be either a diverging or a converging lens.Which statement about a single thin lens is correctA diverging lens always produces a virtual upright image.
13.7 millimeters
13.73076923 mm.
Using the magnification equation m = - v / u. The image should be 13.73 mm in front of the lens
7
13.7 millimeters
13.7 millimetersThis answer is correct, but the formula is most important.The formula is:Hi = height of imageHo = height of objectSi = Distance of image from lensSo = Distance of object from lensYou are trying to find Si, so that is your unknown.Here is your formula: Hi/Ho = Si/SoOr in this case: 3.5/13 = Si/51The rest is basic algebra.Good luck!You can use the ratio equation; (Image Height)/(object height) = - (image location)/(object location) In your case you will get a negative location which means the image is on the same side of the lens as the incoming light.
hi/ho = di/do di = dohi/ho di = (51mm)(3.5mm)/(13mm) di = 14mm * rounded to 2 significant figures The image would be 14mm in front of the lens.
The relation between the distance and height of an object and the image goes like this:L1/H1=L2/H2where L1 and H1 is the Length of the object from the lens and H1 is the height of the object respectively. Same goes for L2, H2 except these are for the image of the same object.If you put values in the above formula, the distance of the image from the lens comes out to be 13.73mm
6 millimeters
Pizza
An object is located 51mm from a diverging lens the object has a height of 13mm and the image height is 3.5mm?Diverging lens do not form real images.When parallel rays of light passes thru a diverging lens, the rays diverge (spread apart) on the other side of the lens. It forms a virtual image. The object will look smaller.The image is on the same side of the lens as the object, so f is negative.Do = 51mm Ho = 13mmDi = ______ Hi = 3.5mmDi = -13.7mm1/Di + 1/Do = 1/f1/-13.5 + 1/51 = 1/ff = -18.36 mm
Using the expression v/u = Image size / object size we can find the value of v. v = 15 * 3.5/13 = 4 (nearly) So approximately at a distance of 4 mm in front of the lens the image is located on the same side of the object.