A converging lens has a focal length of 7.5 cm. An object is
placed 5 cm from the lens.
a. Draw the rays tracing diagram to locate the image. Is the image
real or virtual?
b. Calculate the location and the magnification for this image.
A converging lens has a focal length of 7.5 cm. An object is placed 5 cm...
2. A thin converging lens has a focal length of 10.0 cm. An object is placed 30.0 cm from this lens. Use a sheet of the graph paper provided at the back of this manual to draw a ray diagram that shows the image formed by this lens. Use any two of the three principal (or special) rays and an appropriate scale. Hint: you could let I cm on your ray diagram represent 5 cm of the actual measurements:this scale...
A convex mirror has a focal length of 70 cm. An object is placed 40 cm from the mirror. a. Draw the ray tracing diagram to locate the image. Is the image real or virtual? b. Calculate the image location and the magnification for the image. Compare your results with part a.
A concave mirror has a focal length of 45 cm. An object is placed 35 cm from the mirror. a. Draw the ray tracing diagram to locate the image. Is the image real or virtual? b. Calculate the image location and the magnification for the image. Compare your results with part a. c. What would the image location and magnification be if the object were 55 cm from the mirror?
2. A thin converging lens has a focal length of 10.0 cm. An object is placed 30.0 cm from this lens. Use a sheet of the graph paper provided at the back of this manual to draw a ray diagram that shows the image formed by this lens. Use any two of the three principal (or special) rays and an appropriate scale. Hint: you could let 1 cm on your ray diagram represent 5 cm of the actual measurements:this scale...
A diverging lens has a focal length of 65 cm. It is located 90 cm from an object. a. Draw the ray tracing diagram to locate the image. Is the image real or virtual? b. Calculate the image location and the magnification for the image. Compare your results with part a.
a) Suppose an object is placed 7.50cm from a converging lens with a 5cm focal length. Use ray tracing to get the image and measure image distance. Describe the image: (upright or inverted, real or virtual, bigger or smaller). b) Use the lens equations to calculate the location of the image and its magnification. Describe the image: (upright or inverted, real or virtual, bigger or smaller).
1.) An object is placed in front of a diverging lens with a focal length of 17.7 cm. For each object distance, find the image distance and the magnification. Describe each image. (a) 35.4 cm location _____cm magnification _____ nature real virtual upright inverted (b) 17.7 cm location _____ cm magnification _____ nature real virtual upright inverted (c) 8.85 cm location _____ cm magnification _____ nature real virtual upright inverted 2.) An object is placed in front of a converging lens...
A 4.0 cm tall object is 5.0 cm in front of a diverging lens with a focal length of -6.0 cm. A converging lens with a focal length of 6.0 cm is located 8.0 cm behind the diverging lens. (As viewed from the side, from left to right, the sequence is object - diverging lens - converging lens - observer. Rays then travel from left to right through the system.) (a) Use ray tracing to draw image 1 and image...
(a) An object is placed 7.5cm from a converging lens having a 5cm focal length. Use ray tracing to get the image and measure image distance. Describe the image from your drawing (upright/inverted, real/virtual, bigger/smaller). Please use ruler to get as accurate as you could (b) Use the lens equations to calculate the location of the image and its magnification. Describe the image from your calculation (upright/inverted, real/virtual, bigger/smaller).
Part A: A diverging lens has of focal length of 15.0 cm. An object is placed 21 cm to the left of the lens. a) draw a ray diagram showing the situation. b) find the location of the image produced by the lens (mind the signs). Part B: A converging lens is located 30 cm to the right of the previously mentioned diverging lens (part A). As a result, the image you found in part (a) is now instead located...