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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
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∙ 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?
The formula used to calculate the image distance for a diverging lens is 1/f = 1/d_o + 1/d_i, where f is the focal length of the lens, d_o is the object distance, and d_i is the image distance. Given the object distance of 51 mm, the object height of 13 mm, and the image height of 3.5 mm, the image distance from the lens can be calculated using the equation and appropriate algebraic rearrangements.
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 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?
Since the image height is smaller than the object height, it is a virtual image. Using the thin lens equation (1/f = 1/d_o + 1/d_i), where d_o is the object distance and d_i is the image distance, and assuming a diverging lens, the image distance is found to be -17.17 mm. This means the image is located 17.17 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 image distance can be calculated by the lens formula: 1/f = 1/d_o + 1/d_i, where f is the focal length, d_o is the object distance (negative for diverging lens), and d_i is the image distance. Given d_o = -51 mm, h_o = 13 mm, and h_i = 3.5 mm, you can use the magnification formula: h_i/h_o = -d_i/d_o to find d_i, which is approximately -10.5 mm. So, the image is located 10.5 mm in front of the lens.
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?
Since the image is virtual and upright, it is located on the same side as the object. Using the lens formula 1/f = 1/dO + 1/dI, where f is the focal length, dO is the object distance, and dI is the image distance, you can calculate the image distance. Given the object distance (51 mm), object height (13 mm), and image height (3.5 mm), it would be possible to determine the image distance and thus find out the distance from the lens at which the image is located.
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.
An object is located 51mm from a diverging lens the object has a height of 13mm and the image height is 3.5 how far in front of the lens is the image located?
The magnification formula for a diverging lens is M = -qi/qo = -di/do, where M is the magnification, qi is the image distance, qo is the object distance, di is the image height, and do is the object height. Plugging in the given values, we get 3.5 = -qi/51 => qi = -178.5 mm. Therefore, the image is located 178.5 mm in front of the lens.
