Use the worked example above to help you solve this problem. A diverging lens of focal length f = -9.0 cm forms images of an object situated at various distances. (a) If the object is placed p, = 27.0 cm from the lens, locate the image, state whether it's real or virtual, and find its magnification. cm M = (b) Repeat the problem when the object is at p, = 9.0 cm. cm M = (c) Repeat the problem again when the object is 4.50 cm from the lens. cm M =
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- Two thin lenses with focal lengths of magnitude 15.0 cm, the first diverging and the second converging, are placed 12.00 cm apart. An object 3.00 mm tall is placed 5.50 cm to the left of the first (diverging) lens. Where is the image formed by the first lens located? Please provide a detailed explanation of the process. How far from the object is the final image formed? Please describe the steps taken to reach to your conclusion.= 1.0). An object (A) is Assume that a thin lens (r, placed in the left side of lens (d = 45mm) and the image of the object is located in the right side of the lens at position (B) (d' = 90mm). 1) Compute the refraction index of the lens (n2); = -r2 = 40mm) is placed in the air (n 2) If placing another negative thin lens (f2 = -180mm) immediately behind the first lens to create a doublet lens in which the spacing between two lenses is zero (as shown in the right figure), what is the focal length of this doublet? 3) Where is the final image position (C) of the object (A) through this new doublet? (Note: Assuming this doublet lens is a combined thin lens too without considering its thickness).A double convex lens has radii of 5 cm and 20 cm, a thickness of 2 cm and an index of 1.5. Locate the principal planes and focal points and compute the image distance for an object 16.4 cm in front of V₁. Determine the ffl and bfl.
- Please ASAPUse the worked example above to help you solve this problem. The near point of a patients eye is 50.0 cm. (a) What focal length must a corrective lens have to enable the eye to see clearly an object 29.9 cm away? Neglect the eye-lens distance. 69.0 x cm (b) What is the power of this lens? +1.45 X diopters (c) Repeat part (b), taking into account the fact that, for typical eyeglasses, the corrective lens is 2.00 cm in front of the eye. +2.21 X dioptersHow far from a converging lens with a focal length of 35 cmcm should an object be placed to produce a real image which is the same size as the object? Express your answer to two significant figures and include the appropriate units. dodo = nothingnothing
- An object is placed 30 cm from the center of a 50-cm-focal-length converging lens. (a) Draw a ray diagram and show where the image is located. (Your diagram MUST be legible and correct to earn points). (b) Find the location of the image analytically using the thin lens equation. (c) What is the magnification? (Pay attention to the signs).Needs Complete solution with 100 % accuracy.Use the worked example above to help you solve this problem. The near point of a patients eye is 50.0 cm. (a) What focal length must a corrective lens have to enable the eye to see clearly an object 28.5 cm away? Neglect the eye-lens distance. 50 x cm (b) What is the power of this lens? 2 diopters (c) Repeat part (b), taking into account the fact that, for typical eyeglasses, the corrective lens is 2.00 cm in front of the eye. 2.26 X diopters
- Photographs at a scale of S are required to cover an area X - mi square. The camera has a focal length and focal plane dimensions of 9 * 9in . If endlap is 60% and sidelap 30\% how many photos will be required to cover the area for the data given below? ▾ Part A S = 1/6500 ; X = 8 f = 152.4mn Express your answer as an integer.Find the image distance for a diverging lens of focal length magnitude |f|=10cm for an object located a distance of 13cm in front of the lens. Find the magnification of the image in problem. Describe the image in problem.