object distance= -30cm
image distance= 15cm
magnification= image distance/object distance=- 15/30= - 0.5
image is inverted and real.
2. A thin converging lens has a focal length of 10.0 cm. An object is placed...
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...
2. An object is placed a distance of 25.0 cm from a converging lens having a focal length of 10.0 cm. (a) What is the distance of the image from the lens? (b) What is the magnification of the image? (c) Draw a (somewhat to scale) ray diagram showing the lens, object, and image. (d) Is the image real or virtual?
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 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.
An object is on the left side of a thin converging lens. The object is located at a distance of 6 cm away from a thin converging lens with focal length of 2 cm. Use the thin lens equation (1/f = 1/s' + 1/s) to predict the following: (a) Location of the image? (b) Magnification of the image (including inverted versus non-inverted)? (c) Real or virtual? Draw diagram please!
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).
Problem 1: Ray tracing with a converging lens A 17 cm high object is located 50 cm away from a converging lens with a focal length of 30 cm. The drawing below is to scale (but is not necessarily at a scale of 1:1). A. Draw a ray diagram to find the image, including the height and orientation of the image: you only need to draw 2 of the special rays, but you can draw more if you'd like. Use...
A 4.0-cm tall object is placed 50.0 cm from a diverging lens having a focal length of magnitude 25.0 cm. Draw the ray diagram for this situation. What is the location and height of the image? Is the image real or virtual? • Draw an optical axis with the lens centered on the axis. Represent the object at the correct distance from the lens • Draw the three "special rays" from the top of the object • Extend the rays...
An object is located at a distance of 6 cm from a thin converging lens with focal length of 2 cm. A diverging lens is located 4 cm from the converging lens and 10 cm from the object. The diverging lens has a focal length of -3 cm. Note: To handle a multiple lens system, we treat them independently. We first find the image created by the first lens. We then use the image from the first lens to act...
A converging lens has a focal length of 46.0 cm. If an object is at a distance of 13.1 cm from the lens, determine the location d of the image, the magnification m, and the type of image, if it exists. d = -18.4 cm, m = 1.4, virtual and upright d = +18.4 cm, m = -1.4, real and inverted image does not exist d = 56.2 cm, m = -0.222, real and inverted d = -23.0 cm, m...