An object of height 3 cm is placed at a distance of 25 cm in front of a converging lens of focal length 20 cm, to be referred to as the first lens. Behind the lens there is another converging lens of focal length 20 cm placed 10 cm from the first lens. Find the location, orientation, and size of the final image.
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An object of height 3 cm is placed at a distance of 25 cm in front of a converging lens of focal length 20 cm, to be referred to as the first lens. Behind the lens there is another converging lens of focal length 20 cm placed 10 cm from the first lens. Find the location, orientation, and size of the final image.
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- An object (height = 5.0 cm) and its image are on opposite sides of a converging lens. The object is located 11.0 cm from the lens. The image is located 5.3 cm from the lens. Determine the image height (in cm). Enter the numerical part of your answer to two significant figures. Hint: Remember that the sign of the image height is significant.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 3 cm tall object is placed 16 cm from a converging lens with a focal length of 12 cm. A diverging lens with a focal length of 10 cm is placed 36 cm behind the converging lens. Both lenses have the same principal axis. Draw the ray diagram to graphically find the final image. Need only handwritten solution only (not typed one).
- An object is placed 25 cm in front of a converging lens of focal length 20 cm. 30 cm past the first lens is a second diverging lens of magnitude focal length 25 cm. What are the resulting image position relative to second lens and total magnification of the object in this setup?Is the final image real or virtual? Is the final image upright or inverted?A 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.A converging lens of focal length 20 cm is placed 30 cm in front of another converging lens of focal length 4.0 cm. An object is placed 100 cm in front of the first lens. Determine(a) the location of the final image, (b) its orientation, and (c)whether it is real or virtual.The lens-maker’s equation for a lens with index n1 immersed in a medium with index n2 takes the form A thin diverging glass (index = 1.50) lens with R1 = −3.00 m and R2 = −6.00 m is surrounded by air. An arrow is placed 10.0 m to the left of the lens. (a) Determine the position of the image. Repeat part (a) with the arrow and lens immersed in (b) water (index = 1.33) (c) a medium with an index of refraction of 2.00. (d) How can a lens that is diverging in air be changed into a converging lens?
- A 3 cm tall object is placed 2 cm to the left of a converging lens that has a focal length with a magnitude of 4 cm. A diverging lens with a focal length of magnitude 8 cm is placed 10 cm to the right of the first lens. What is the magnification of the final image produced by these two lenses? Make sure you say whether it is bigger/smaller and upright/inverted.A converging lens of focal length f=2 cm is used to focus the image of an object onto a screen. The object and screen are separated by 15 cm. The lens is placed between the object and the screen, at a distance from the screen such that the image of the object is focused into the screen. At what distance from the screen must the lens be placed in order to have an image magnification < 1? Provide your answer to two significant figures. distance = cm.There is a lens combinations between two converging lenses that have focal lengths of f1=13 cm (lens 1) and f2=16 cm (lens 2). They are separated by 56 cm along a horizontal line. There is a lit candle (height ho) placed 36 cm in front of the lens 1 (f1=13 cm). Thus, the separation between the candler and the second lens (f2=16 cm) is 92 cm Please calculate actual postion of final image. Is it real/imaginary?