A diverging lens (f=-12.0 cm) is located 22.0 cm to the left of a converging lens (f=34.0 cm). A 3.70 cm tall object stands to the left of the diverging lens, exactly at its focal point. (a) Determine the distance of the final image relative to the converging lens. (b) What is the height of the final image (including the proper algebraic sign)?
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A diverging lens (f=-12.0 cm) is located 22.0 cm to the left of a converging lens (f=34.0 cm). A 3.70 cm tall object stands to the left of the diverging lens, exactly at its focal point. (a) Determine the distance of the final image relative to the converging lens. (b) What is the height of the final image (including the proper algebraic sign)?

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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.The focal length of a diverging lens is negative. If f = −16 cm for a particular diverging lens, where will the image be formed of an object located 45 cm to the left of the lens on the optical axis? cm to the left of the lensWhat is the magnification of the image?an object of height 5 cm is placed 20 cm in front of a converging lens at focal length 10 cm. Behind the converging lens, and 25cm from it, there is a diverging lens of the focal length of 6 cm. Find the location of the final image, in centimeters, with respect to teh diverging lens. what is the magnification of the final image? what is the height of the final image?
- (a) A 1.00 cm-high object is placed 4.35 cm to the left of a converging lens of focal length 7.85 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. cm---Select--- position height Is the image inverted or upright? upright inverted Is the image real or virtual? real virtual cm (b) What If? If an image of opposite characteristic, i.e., virtual if the image in part (a) is real and real if the image in part (a) is virtual, is to be obtained, what is the minimum distance (in cm), and in which direction, that the object must be moved from its original position? cm distance direction ---Select---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
- 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.An object of 1 cm tall is placed 3cm in front of a converging lens of focal length of 2 cm. (a) Use ray tracing to find the image. (b) Use the lens equation to find the image distance and compare it to your ray tracing and find the percentage error.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 diverging lens (f1 = −11.5 cm) is located 20.0 cm to the left of a converging lens (f2 = 21.5 cm). A 4.0-cm-tall object stands to the left of the diverging lens, exactly at its focal point. What is the height of the final image (including proper algebraic sign)?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 system of two lenses forms an image of an arrow at x = x3 = 57.4 cm. The first lens is a diverging lens located at x = 0 and has a focal length of magnitude f₁ = 11.5 cm. The second lens is located at x = x₂ = 25.4 cm and has an unknown focal length. The tip of the object arrow is located at (x,y) = (xo, Yo) = (-36 cm, 20.6 cm). (x,y) 1) What is x₁, the x-coordinate of image of the arrow formed by the first lens? cm Submit 2) What is y₁, the y-coordinate of the image of the tip of the arrow formed by the first lens? cm Submit X₂ Real and Inverted Real and Upright Virtual and Inverted Virtual and Upright cm Submit (+) 3) What is f2, the focal length of the second lens. If the lens is a converging lens, f2 is positive. If the lens is a diverging lens, f2 is negative. cm Submit + 4) What is y3, the y-coordinate of the image of the tip of the arrow formed by the two lens system? (+ 5) The positions of the two lenses are now interchnaged (i.e., the second lens is moved to x = 0 and the…