Spherical mirrors. Object O stands on the central axis of a spherical mirror. For this situation object distance is p, - +16 cm, the type ofmirror is concave, and then the distance between the focal point and the mirror is 12 cm (without proper sign). Find (a) the radius ofcurvaturer (including sign). (b) the image distance i, and (c) the lateral magnification m. Also, determine whether the image is (d) real orvirtual, (e) inverted from object O or noninverted, and (f) on the same side of the mirror as O or on the opposite side.(a) NumberUnits(b) NumberiUnits(c) NumberiUnits(d)(e)(f)

Given data p=16 cmf=12 cm Here, P is the object distance. While f is the focal distance. First of…

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Spherical mirrors. Object O stands on the central axis of a spherical mirror. For this situation object distance is ps = +11 centimeters, the type of mirror is convex, and then the distance between the focal point and the mirror is 8.2 cm (without proper sign). Find (a) the radius of curvature r (including sign), (b) the image distance i, and (c) the lateral magnification m. Also, determine whether the image is (d) real or virtual, (e) inverted from object O or noninverted, and (f) on the same side of the mirror as O or on the opposite side. (a) Number (b) Number (c) (d) (e) (f) Number Hi > > Units Units Units >
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a) An object is placed 40.0cm in front of a diverging lens with a focal length of 19.2 cm. Please provide your answers in units of cm if necessary. Find the image distance. I answered this part and I got -13 as an answer. But i can't solve this second part: b) Suppose the object is brought to a distance of 11.7 cm in front of the lens. Where is the image now with respect to its previous location? (Note: Ensure the sign convention you use is consistent by treating all image distances on the object side of the lens as negative.) c)How has the height of the image changed if the object is 80.7 cm tall? plz help me answer it. thanks
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Can you please explain how...well...how this works. Can you draw a diagram or something? It seems like if you use m=v/u and enter 25 cm for v, then you're saying that the image is 25cm from whatever the rest of the system is in the equation. m is the magnification of the lens, so when you arrive at the value for u, that seems like it should be saying that if you put the iron 8.3 cm from the lens, then it will project the image 25 cm from the lens. Not 25 cm in front of your eyes. I don't understand how the answer ends up applying to the distance to the eyes. Is there some algebra that needs to be done to link the distance from the image to the distance to the eyes? By the way, what do v and u stand for? I know they're distances, but what words do the letters represent?

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