5) How large is the image of a 2.00-mm diamond when viewed through a magnifying glass of focal distance 4.00 cm when the diamond is held at 3.00 cm away from the lens? Include a scaled ray diagram.
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- A 1.8 cm-tall object stands in front of a converging lens. It is desired that a virtual image 2.7 times larger than the object be formed by the lens. How far from the lens (in cm) must the object be placed to accomplish this task, if the final image is located 18 cm from the lens? cme) The object is between F and optical center Image: 1. 2. 3. Principal axisAn object of 1 cm tall is placed in front of a converging lens of focal length of 2 cm. The object is 2.5 cm away from the mirror. (a 5') Use the ray tracing to find the image. (b 5') Compute the image distance and height using the lens equation.pls answer a-b pls
- Determine the image distance and size for a 10.00-cm tall object placed a) 60.0 cmb) 10 cm from a concave mirror having a focal length of 30.0 cm. Draw light ray diagrams for each case.A converging lens has a focal length of 10.9 cm. (a) Locate the object if a real image is located at a distance from the lens of 32.7 cm. X distance Your response differs from the correct answer by more than 100%. cm ---Select--- location (b) Locate the object if a real image is located at a distance from the lens of 43.6 cm. distance cm location ---Select--- (c) Locate the object if a virtual image is located at a distance from the lens of -32.7 cm. distance cm location ---Select--- (d) Locate the object if a virtual image is located at a distance from the lens of -43.6 cm. distance cm location ---Select--- Need Help? Read It Submit Answer4-
- The near point of a person's eye is 72.8 cm. (Neglect the distance from the lens to the eye.) (a) To see objects clearly at a distance of 24.0 cm, what should be the focal length of the appropriate corrective lens? cm (b) To see objects clearly at a distance of 24.0 cm, what should be the power of the appropriate corrective lens? dioptersVARIANT 23 Suppose an object such as a book page is held 15 cm from a concave lens of focal length - 10.0 cm. Such a lens could be used in eyeglasses to correct pronounced nearsightedness. (a) Using the thin lens equations to calculate the location of the image. (b) What magnification is produced? (c) Use ray tracing to get an approximate location for the image.(10%) Problem 8: A candle (h, = 0.39 m) is placed to the left of a diverging lens (f = -0.051 m). The candle is d, = 0.38 m to the left of the lens. A 25% Part (a) Write an expression for the image distance, d;. A 25% Part (b) Numerically, what is the image distance in meters? V 25% Part (c) Is this real or virtual? Virtual v Correct! D A 25% Part (d) Numerically, what is the image height, h; in meters?
- 6) An object is placed very far in front of a converging lens of focal length 13.2 cm. Find the location of the image, be sure to indicate which side of the lens the image is on.A 1.9 cm-tall object stands in front of a converging lens. It is desired that a virtual image 2.1 times larger than the object be formed by the lens. How far from the lens (in cm) must the object be placed to accomplish this task, if the final image is located 16 cm from the lens? cmA 1.00-cm-high object is placed 3.95 cm to the left of a converging lens of focal length 8.30 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 height cm cm -Select- -Select- Is the image real or virtual? O real O virtual in front of the second lens behind the second lens Is the image inverted or upright? O upright O inverted 4