A converging lens (f = 10 cm) and a second converging lens (f = 30 cm) are placed 25 cm apart, and an object is placed 30 cm in front of the first converging lens. Object's height is 2 cm. Where is the location of the final image?
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- A 2.5 cm-tall object stands in front of a converging lens. It is desired that a virtual image 2.6 times larger than the object be formed by the lens. How far from the lens (in cm) must the object be placed to accomplish this task, if the final image is located 13 cm from the lens? cm?An object with a height of -0.040 m points below the principal axis (it is inverted) and is 0.200 m in front of a diverging lens. The focal length of the lens is -0.27 m. (Include the sign of the value in your answers.) (a) What is the magnification? (b) What is the image height? m (c) What is the image distance? mThe focal length of a diverging lens is negative. If f= -26 cm for a particular diverging lens, where will the image be formed of an object located 53 cm to the left of the lens on the optical axis? cm to the left of the lens What is the magnification of the image?
- A 1 [cm] tall object is 50 [cm] away from a converging lens of focal length f = 30 [cm]. A second converging lens of focal length f = 80 [cm] is placed 2 [m] behind the first lens. a) Where is the image formed by the first lens (ignoring the second lens)? How big is it? b) Now, this image is the object for the second lens. Where is the second (final) image formed by the second lens? How big is it?An object with a height of -0.040 m points below the principal axis (it is inverted) and is 0.140 m in front of a diverging lens. The focal length of the lens is -0.25 m. (Include the sign of the value in your answers.) (a) What is the magnification? (b) What is the image height? m (c) What is the image distance? mA 1.00-cm-high object is placed 3.95 cm to the left of a converging lens of focal length 7.80 cm. A diverging lens of focal length –16.00 cm is 6.00 cm to the right of the converging lens. Find the position and height of the final image. position 7.5 cm in front of the second lens v 0.5466 height Calculate the magnification produced by each lens. Then consider how the magnification relates image size and object size for each lens to find the height of the final image. cm Is the image inverted or upright? upright O inverted Is the image real or virtual? real virtual
- An object is located 50.0 m to the left a converging lens of focal length 15.0 cm. A diverging lens of focal length 4.20 cm is placed 10.0 cm to the right of the converging lens. Locate (final image position) and describe (real or virtual, upright or inverted, reduced or enlarged) the final image of the object formed by the two-lens system.I continue to get the wrong answer. TIA.