What would be the image distance for the +20.0 cm lens if the object distance were +60.0 cm? What would be the corresponding magnification of the image? Would the image be upright or inverted
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What would be the image distance for the +20.0 cm lens if the object distance were +60.0 cm? What would be the corresponding magnification of the image? Would the image be upright or inverted?
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- A microscope has an objective lens with a focal length of 16.22 mm and an eyepiece with a focal length of 9.86 mm. With the length of the barrel set at 34.0 cm, the diameter of a red blood cell's image subtends an angle of 1.43 mrad with the eye. If the final image distance is 34.0 cm from the eyepiece, what is the actual diameter of the red blood cell? Hint: To solve this equation, go back to basics and use the thin - lens equation.A converging lens (n = 1.50) has radii of curvature of 2.50 cm and 5.00 cm; you will need to assign the correct signs. Determine the focal length of lens-1. A cup of Darjeeling (10 cm tall) is placed 5.00 cm in front of lens-1. Compute the position and magnification of the image. A diverging mirror, of radius 15 cm, is placed 20 cm to the right of the converging lens. Determine the (i) position, (ii) final magnification (and its image characteristics), and (iii) size of the final image. Draw a good ray diagram showing the location of the first and final image.The near point of a person's eye is 60.0 cm. What is the approximate power of the correcting lens that allows the person to clearly see objects placed 30.0 cm away? Neglect the eye-lens distance. O (A) -5.00 diopters O (B) -1.67 diopters C (C) +1.67 diopters O (D) +5.00 diopters Need Help? Read It
- Two converging lenses having focal lengths of f1 = 11.6 cm and f2 = 20.0 cm are placed d = 50.0 cm apart, as shown in the figure below. The final image is to be located between the lenses, at the position x = 30.9 cm indicated./ (a) How far to the left of the first lens should the object be positioned in cm? (b) What is the overall magnification of the system?What should be the object distance from the convex lens (f=15 cm), that makes a virtual image that is 3.0 times magnified? O 33.3 cm 10 cm 4.8 cm O 5.4 cmA converging lens with a focal length of 40 cm and a diverging lens with a focal length of -40 cm are 150 cm apart. A 3.0-cm-tall object is 60 cm in front of the converging lens. Calculate the image height. Express your answer in centimeters to two significant figures. h = IVE ΑΣΦ ? cm
- In a compound microscope, the object is at 15 mm left from the objective lens. The lenses are 300 mm apart. First and the final image are formed at a distance of 100 cm and 25 mm right of the respective lenses. The ratio of two focal lengths is a/b. What is the value of a+b?.A microscope has an objective lens with a focal length of 16.22 mm and an eyepiece with a focal length of 9.92 mm. With the length of the barrel set at 34.0 cm, the diameter of a red blood cell's image subtends an angle of 1.43 mrad with the eye. If the final image distance is 34.0 cm from the eyepiece, what is the actual diameter of the red blood cell? Hint: To solve this equation, go back to basics and use the thin-lens equation.A certain lens has a focal length of +20cm. How far from the lens must an object be placed so that it produces an image with a magnification of -1.37? Is the image real or virtual?
- A certain slide projector has a 135 mm focal length lens. (a) How far away is the screen (in m), if a slide is placed 139 mm from the lens and produces a sharp image? m (b) If the slide is 18.0 by 45.0 mm, what are the dimensions of the image? (Enter your answers from smallest to largest in cm.) cm by cmCan you please explain how...well...how this works. Can you draw a diagram or something? It seems like if you use m=v/u and enter 25 cm for v, then you're saying that the image is 25cm from whatever the rest of the system is in the equation. m is the magnification of the lens, so when you arrive at the value for u, that seems like it should be saying that if you put the iron 8.3 cm from the lens, then it will project the image 25 cm from the lens. Not 25 cm in front of your eyes. I don't understand how the answer ends up applying to the distance to the eyes. Is there some algebra that needs to be done to link the distance from the image to the distance to the eyes? By the way, what do v and u stand for? I know they're distances, but what words do the letters represent?Need help:. You have an object 16.0 cm away from a diverging lens of focal length f = –16.0 cm. From this information, determine the magnification of the lens. (hint: work backwards from magnification to determine the approach to this problem)