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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?
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∙ 14y ago
Wiki User
∙ 14y ago
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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 of the lens is?
13.7 millimeters
If an object 18mm high is placed 12mm from a diverging lens and the image is formed 4mm in front of the len what is the height of the image?
6mm
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?
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.
What is the height of the shadow when the distance from the object to the screen is 60 cm ............................?
More information, or an image, is required to answer this question.
What is the height of Nishas image if she stood 1.6m away from a long plane mirror actually her height is 165cm tall?
The image formed in a plane mirror is virtual, and equal in size to the object. so the size of Nishas image will also be 165cm, equal to her own size
An object 51 mililmeters from a diverging lens the object has a height of 13 milimeters and the image height is 3.5 milimeters how far in front of the lens is image located?
Pizza
An object 51 millimeters from diverging lensthe object has a height of 13 millimeters and the image height is 3.5 millimetershow far in frontof the lens is the image located?
13.73076923 mm.
An object is 51 millimeters from a diverging lens the object has a height of 13 millimeters and then image height is 3.5 millimeters How far in front of the lens is the image located?
13.7 millimeters
An object is located 51 millimeter from a diverging lens the object has a height of 13 millimeter and the image heightg is 3.5 millimeters how far in front of the lens is the image located?
7
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 of the lens is?
13.7 millimeters
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
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.
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
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.
If an object 18mm high is placed 12mm from a diverging lens and the image is formed 4mm in front of the len what is the height of the image?
6mm
An object located 51 millimeters from a diverging lens the object is 13 millimeters high and the image 3.5 millimeters how far in front of the lens is the image located?
13.7 millimeters
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?
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.
