An object with a height of -.06m points below the principal axis (it is inverted
ID: 1836072 • Letter: A
Question
An object with a height of -.06m points below the principal axis (it is inverted) and is 0.160 m in front of a diverging lens. The focal length of the lens is 0.31 m. (Include the sign of the value in your answers.) (a) What is the magnification?(b) What is the image height?
m
(c) What is the image distance?
m An object with a height of -.06m points below the principal axis (it is inverted) and is 0.160 m in front of a diverging lens. The focal length of the lens is 0.31 m. (Include the sign of the value in your answers.) (a) What is the magnification?
(b) What is the image height?
m
(c) What is the image distance?
m 0.31 (a) What is the magnification?
(b) What is the image height?
m
(c) What is the image distance?
m
Explanation / Answer
object distance u = 0.16 m
focus f = - 0.31 m
1/u + 1/v = 1/f
1/0.16 + 1/v = -1/0.31
1/v = -1/0.16 - 1/0.31
image distance v = - 0.105 m this shows image is virtual and on same side as object
magnification = - v/u = 0.66
image height = magnification * object height = - 0.04 m that means image is inverted too
answers a) 0.66 b) -0.04 c) - 0.105 m