given
object distance d0 = 50 cm
if the image is formed on the same side of the object
so it is virtual image and upright
if the object distance is less than the focal length of the lens
then image is virtual, upright
lens is a converging
answer is
upright,converging, negative and virtual
An object is placed 50.0cm in front of a lens. The image forms on the same...
An object is placed 50.0cm in front of a lens. The image forms on the same side of the lens and is larger than the object. The image is (upright or inverted), the lens is (converging/diverging), image distance is (positive/negative), the image is (real, virtual) O inverted, diverging, positive, real upright, converging, positive, virtual inverted, converging, positive, virtual upright, converging, positive, real upright, converging, negative, virtual upright, converging, negative, real inverted, converging, positive, real O inverted, diverging, negative, real
Question 18 (8 points) An object is placed 50.0cm in front of a lens. The image forms on the same side of the lens and is larger than the object. The image is (upright or inverted), the lens is (converging/diverging), image distance is (positive/negative), the image is (real, virtual) upright, converging, negative, real inverted, converging, positive, virtual upright, converging, negative, virtual upright, converging, positive, virtual upright, converging, positive, real O inverted, converging, positive, real inverted, diverging, positive, real inverted, diverging,...
A charge, q=91.0000 microCoulombs on a particle with mass m=1.00000 milli- grams, moves through a pipe from the origin to a point at coordinate x=1.40000m and y=1.8000m. All space is filled with a uniform electric field E=1,900.00000N/C and pointing parallel to the x axis. What is the change in electric potential as the mass moves from initial to final positions (in VOLTS) An object is placed 50.0cm in front of a lens. The image forms on the same side of...
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...
Please help with this 4 part question Light with wavelength = 700.0000nm is incident on a double slit with spacing d=50.00000micrometers. The distance to the screed is 0.5000meters. What is the spacing between neighboring constructive fringes near the center of the screen? (in meters) Light with wavelength 500nm is incident upon a surface at an angle of 40.0 degrees, refracts and enters a second medium. The light that enters the second medium travels faster than in the first medium. Compared...
An object is placed to the right of a converging lens and an image is formed to the left of the lens. lens object image 1) Following the standard sign convention for the lens equation 5 + 5 = }, the image distance is: positive negative Submit 2) The image is: real virtual Submit 3) For this lens, the sign of the focal length is: positive negative Submit Help 4) An object is placed to the left of a converging...
a small object is placed 25.0 cm from a converging lens of focal length 40.0cm.the object is to the left of the lens where is the image? same as the previous question. which statement is true about the image? It is real, upright and larger than the object It is virtual, upright and larger than the object. It is virtual, inverted and larger than the object. It is virtual, inverted and smaller than the object. It is real, inverted and...
A small object is placed 25 cm to the left of a converging lens of focal length 40 (same as question 11). Which statement is true about the image? It is real, inverted and larger than the object. It is real, inverted and smaller than the object. It is virtual, inverted and larger than the object. It is virtual, inverted and smaller than the object. It is virtual, upright and larger than the object. It is real, upright and larger...
A 2.0 cm tall object is placed 10.0 cm in front of converging lens of focal length 8 cm. What can you say about the image formed by the lens? de the image is real, magnified and inverted the image is virtual, diminished and inverted O the image is virtual, magnified and upright the image is real, magnified and upright the image is real, diminished and inverted
An object is placed 7 cm in front of a converging lens. An image forms which is inverted and has a magnification of -2. Where should the object be moved such that the image is upright and has a magnification of 4? Answer in cm.