An object of height 1 cm is placed 2 cm in front of a diverging lens of focal length of 2 cm. (a,5) Use the ray tracing method to locate the image and measure your image distance. (b,5) Then use the lens equation to find the theoretical value of the image distance. (c,5) What is your percentage error in the image distance?
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An object of height 1 cm is placed 2 cm in front of a diverging lens of focal length of 2 cm.
(a,5) Use the ray tracing method to locate the image and measure your image distance.
(b,5) Then use the lens equation to find the theoretical value of the image distance.
(c,5) What is your percentage error in the image distance?
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- 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?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? m
- An object 12 mm size is located 20 cm in front of a converging lens of 15 cm focal length. (a) Find the image distance and magnification. Is the image real or virtual? (b) What is the size of image and will the image be upright or inverted?A microscope objective with a magnification of 20 is used in a microscope with a magnification of 56. Find the focal length of the ocular if the ocular's image is formed at the eye's near point, 27 cm from the eye. Neglect the distance between the eye and the ocular.Two converging lenses, both with focal lengths 34.9 cm are placed a distance 78.9 cm apart. An object is placed 87.5 cm from the 1st lens. Where is the final image located relative to the 2nd lens? Let + be the back side and - be the front side regarding this distance from the 2nd lens. Measure this in cm.
- 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.An object is placed in front of a diverging lens whose focal length is - 30.7 cm. A virtual image is formed whose height is 0.532 times the object height. How far is the object from the lens?An object with a height of -0.04 m points below the principal axis (it is inverted) and is 0.130 m in front of a diverging lens. The focal length of the lens is -0.22 m. (Include the sign of the value in your answers.) (a) What is the image distance? (b) What is the magnification? (c) What is the image height?
- A lens of focal length 42 mm is used as a magnifier. The object being viewed is 6.8 mm long, and is positioned at the focal point of the lens. The lens is moved closer to the object, so that the image is now 15 cm from the lens. The distance the lens has been moved, in cm, is closest to:An object is placed a distance p to the left of a diverging lens of focal length f1. A converging lens of focal length f2 is placed a distance d to the right of the diverging lens. Find the distance d so that the final image is infinitely far away to the right.An object is 20.0 cm to the left of the convex lens which is 50.0 cm from a concave lens as shown below. The focal lengths of the convex and concave lenses are 10.0 cm and -12.5 cm. respectively. (a) Use the thin lens equation to determine the location of the image made by the convex lens. (hi Treat the image made by the convex lens as the object of the concave lens and determine the location of the final image. (c) Characterize the final image completely. (Reduced or enlarged, real or virtual, Inverted or upright).