In the situation above the converging lens has a focal length of 50 cm, and the
ID: 1520988 • Letter: I
Question
In the situation above the converging lens has a focal length of 50 cm, and the diverging mirror has a focal length of -80 cm. What is the location of the image after it passes through the lens? What is the magnification of the image? Is it upright or inverted? Now the lens is removed and the light from the object strikes the mirror. What is the location of the image after the light reflects off the mirror? What is the magnification of the image? Is it upright or inverted? State and briefly describe each of Maxwell's equations. What is the significance of the equations?Explanation / Answer
B )
i)
for the lens :
f = 50 cm
do = object distance = 100 cm
di = image distance
using the lens equation
1/di + 1/do = 1/f
1/di + 1/100 = 1/50
di = 100 cm
the image is formed at the mirror
ii)
m = - di/do = - 100 / 100 = -1
the image is inverted
iii)
for the mirror :
f = - 80 cm
do = object distance = 200 cm
di = image distance
using the lens equation
1/di + 1/do = 1/f
1/di + 1/200 = 1/(-80)
di = - 57.14 cm located to the right of mirror
iv)
m = - di/do = - (-57.14) / 200 = 0.286
the image is upright