A beam of parallel rays spreads out after passing through a thin diverging lens, as if the rays all came from a point 20.0 cm from the center of the lens. You want to use this lens to form an erect, virtual image that is 1 3 the height of the object. (a) Where should the object be placed? Where will the image be? (b) Draw a principal-ray diagram.
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A beam of parallel rays spreads out after passing through a thin diverging lens, as if the rays all came from a point 20.0 cm from the center of the lens. You want to use this lens to form an erect, virtual image that is 1 3 the height of the object. (a) Where should the object be placed? Where will the image be? (b) Draw a principal-ray diagram.
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- An object with a height of -0.040 m points below the principal axis (it is inverted) and is 0.200 m in front of a diverging lens. The focal length of the lens is -0.27 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= 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).Mehlo
- In the figure, a real inverted image I of an object O is formed by a certain lens (not shown); the object-image separation is d = 57.8 cm, measured along the central axis of the lens. The image is just 1/4 the size of the object. (a) What kind of lens must be used to produce this image? (b) How far from the object must the lens be placed? (c) What is the focal length of the lens? Lens here Axis (a) (b) Number i Units (c) Number i UnitsA 2.0 cm tall object is 40 cm in front of a converging lens that has a 20 cm focal length. Use ray tracing to find the position and height of the image, using a ruler or paper with a grid. Calculate the image position and height and compare with your answer to (a). Is the image real or virtual?It is your first day at work as a summer intern at an optics company. Your supervisor hands you a diverging lens and asks you to measure its focal length. You know that with a converging lens, you can measure the focal length by placing an object a distance ss to the left of the lens, far enough from the lens for the image to be real, and viewing the image on a screen that is to the right of the lens. By adjusting the position of the screen until the image is in sharp focus, you can determine the image distance s′s′ and then use the equation 1s+1s′=1f1s+1s′=1f, to calculate the focal length ff of the lens. But this procedure won't work with a diverging lens−−by itself, a diverging lens produces only virtual images, which can't be projected onto a screen. Therefore, to determine the focal length of a diverging lens, you do the following: First you take a converging lens and measure that, for an object 20.0 cmcm to the left of the lens, the image is 29.7 cmcm to the right of the lens.…
- 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 virtualFor safety reasons, you install a rear-window lens with a -0.299 m focal length in your van. Before putting the van in reverse, you look through the lens and see the image of a person who appears to be 0.339 m tall and 0.243 m behind the van. Determine the following. (a) actual distance of the person behind the van m (b) height of the person
- 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 has a focal length of 20.0 cm. Construct accurate ray diagrams for object distances of (i) 60.0 cm and (ii) 6.67 cm. (a) From your ray diagrams, determine the location of each image. Image (i) distance = Image (ii) distance =