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∙ 14y agoThe 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.
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.
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.
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.
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13.7 millimeters
13.73076923 mm.
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
7
about 2 inches
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.
2.5 feet or 760 milimeters
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.
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.
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.
6mm