An object of height yo = +2 m is placed in front of a concave spherical mirror (
ID: 1520227 • Letter: A
Question
An object of height yo = +2 m is placed in front of a concave spherical mirror (focal length of size 1 m), at the various locations listed in parts a-c.
a.) When the object is placed 3 m away from the mirror, please find:
mage distance =_____ m
type/orientation of image =__real & upright, real & inverted, virtual & upright, virtual & inverted___
magnification = _____
height of image = _____m
b.) When the object is placed 1.5 m away from the mirror, please find:
mage distance =_____ m
type/orientation of image =__real & upright, real & inverted, virtual & upright, virtual & inverted___
magnification = _____
height of image = _____m
c.) When the object is placed 0.5 m away from the mirror, please find:
mage distance =_____ m
type/orientation of image =__real & upright, real & inverted, virtual & upright, virtual & inverted___
magnification = _____
height of image = _____m
The same object (of height = 2 m) is placed in front of a converging lens (focal length of size 1 m), at the various locations listed in parts d-f.
d.) When the object is placed 3 m away from the lens, please find:
mage distance =_____ m
type/orientation of image =__real & upright, real & inverted, virtual & upright, virtual & inverted___
magnification = _____
height of image = _____m
e.) When the object is placed 1.5 m away from the lens, please find:
mage distance =_____ m
type/orientation of image =__real & upright, real & inverted, virtual & upright, virtual & inverted___
magnification = _____
height of image = _____m
f.) When the object is placed 0.5 m away from the lens, please find:
mage distance =_____ m
type/orientation of image =__real & upright, real & inverted, virtual & upright, virtual & inverted___
magnification = _____
height of image = _____m
Explanation / Answer
(a) do = 3 m , f = 1m
we know that the image distance is given by
di = dof / (d0-f) = 3*1 /(3-1) = 3/2 = 1.5 m
Hence the image formed will be at the 1.5 m in front of the mirror .
Image will be real and inverted.
Magnification (m)= -di/do = -1.5/3 = -1/2 =-0.5
we know that m = Image height /object height
image height = object height*magnification = 2*(-0.5) = -1 m
negative sign indicate that the image is inverted.
(b) when do = 1.5 m , therefore
di = dof / (d0-f) = (1.5)*1/(1.5-1) = 3m
Magnification (m) = -di/do = -3/1.5 = -2
Hence image is real and inverted
Image height = Yo*m = 2*(-2) = -4 m
(c) When do = 0.5 m
di = dof / (d0-f) = 0.5*1/(0.5-1) = -1 m
Hence the image will be erect and virtual .
Magnification (m) = -(di/do) = -(-1/0.5) = 2
Image height = Yo*m = 2*2 = 4 m .
(d) we have do = 3 m , f = 1m (since lens is converging therfore focal length should be positive)
di = dof / (d0-f) = 3*(1)/(3-1) = 3/2 = 1.5 m
hence Image will be real and inverted.
Magnification (m) = -(di/do) = -(1.5/3) = -1/2 = -0.5 m
Image height = Yo*m = 2*(-0.5) = -1 m
Negative sign indicate that image is inverted.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.