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# An object is located 51 millimeters from a diverging lens the object has a height og 13 millimeters and the image height is 3.5 millimeters how far in front of the lens is image located?

Updated: 9/17/2023

Wiki User

13y ago

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.

Wiki User

13y ago

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Q: An object is located 51 millimeters from a diverging lens the object has a height og 13 millimeters and the image height is 3.5 millimeters how far in front of the lens is image located?
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### An object is located 51 millimeters from a diverging lens the object has a height of 13 millimeters and the image height is 3.5 millimeters how far in front of the lens is image located?

Using the magnification equation m = - v / u. The image should be 13.73 mm in front of the lens

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### An object is located 51 millimeters from a diverging lens the object has a height of 13 millimeters and the image is 3.5 meters how far in front of the lens is the image located?

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.

### An object is located 51 millimeters from a diverging lensthe object has a height of 13 millimeters and thye image height is 3.5 millimetershow far in front of thelens is the image located?

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

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### An object is located 51mm from a diverging lens the object has a height of 13mm and the image height is 3.5mm?

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

### An object is located 15millimeters from a diverging lens the object has a height of 13 mm and the image height is 3.5 mm how far in front of the lens is the image located?

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.