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A diverging lens located in the y-z plane at x = 0 forms an image of an arrow at x = x2 = -26 cm. The image of the tip of the arrow is located at y = y2 = 4.7 cm. The magnitude of the focal length of the diverging lens is 48.2 cm. light 1) What is X₁, the x-coordinate of the object arrow?.

A diverging lens located in the y-z plane at x = 0 forms an image of an arrow at x = x2 = -26 cm. The image of the tip of the arrow is located at y = y2 = 4.7 cm. The magnitude of the focal length of the diverging lens is 48.2 cm. light 1) What is X₁, the x-coordinate of the object arrow?.

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A diverging lens located in the y-z plane at x = 0 forms an image of an arrow at x = x2 = -26 cm. The image of the tip of the arrow is located at y = y2 = 4.7 cm. The magnitude of the focal length of the diverging lens is 48.2 cm. 1) What is x₁, the x-coordinate of the object arrow?. cm Submit cm Submit 3) A converging lens of focal length fconverging = 15.49 cm is now inserted at x = x3 = -35 cm. In the absence of the diverging lens, at what x co-ordinate, x4, would the image of the arrow form? 2) What is y₁, the y-coordinate of the tip of the object arrow? cm Submit Virtual and upright Virtual and inverted light light (X2,3₂) cm Submit image X3 I Ay If this description seems unclear to you, you can check out the more detailed description of multiple lens systems given in the Optical Instruments unit. What is x5, the x co-ordinate of the final image of the combined system? 5) Is the final image of the arrow real or virtual? Is it upright or inverted? Real and upright Real and inverted X image in absence. of diverging lens X (+) 4) To determine the image of the arrow from the combined converging + diverging lens system, we take the image from the converging lens (in the absence of the diverging lens) to be the object for the diverging lens. If this image is downstream of the diverging lens, the object for the diverging lens is virtual. All this means is that the rays entering the diverging lens are converging towards a location downstream of the diverging lens. The final image can be calculated using this virtual object distance and the focal length of the diverging lens. + + (+)
Transcribed Image Text:A diverging lens located in the y-z plane at x = 0 forms an image of an arrow at x = x2 = -26 cm. The image of the tip of the arrow is located at y = y2 = 4.7 cm. The magnitude of the focal length of the diverging lens is 48.2 cm. 1) What is x₁, the x-coordinate of the object arrow?. cm Submit cm Submit 3) A converging lens of focal length fconverging = 15.49 cm is now inserted at x = x3 = -35 cm. In the absence of the diverging lens, at what x co-ordinate, x4, would the image of the arrow form? 2) What is y₁, the y-coordinate of the tip of the object arrow? cm Submit Virtual and upright Virtual and inverted light light (X2,3₂) cm Submit image X3 I Ay If this description seems unclear to you, you can check out the more detailed description of multiple lens systems given in the Optical Instruments unit. What is x5, the x co-ordinate of the final image of the combined system? 5) Is the final image of the arrow real or virtual? Is it upright or inverted? Real and upright Real and inverted X image in absence. of diverging lens X (+) 4) To determine the image of the arrow from the combined converging + diverging lens system, we take the image from the converging lens (in the absence of the diverging lens) to be the object for the diverging lens. If this image is downstream of the diverging lens, the object for the diverging lens is virtual. All this means is that the rays entering the diverging lens are converging towards a location downstream of the diverging lens. The final image can be calculated using this virtual object distance and the focal length of the diverging lens. + + (+)
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