Consider a spherical mirror and lens separated by 45 cm. The mirror is on the le
ID: 1455150 • Letter: C
Question
Consider a spherical mirror and lens separated by 45 cm. The mirror is on the left with a focal length of 100 cm. The lens is on the right with a focal length of 20 cm. A 5 cm tall object is placed 20 cm to the left of the lens.
a) If you only consider the rays that move to the right from the object, fully characterize the final image in the system. In other words provide final image location, real/virtual, upright/inverted, final magnification, and final image height.
b) If you only consider the rays that move to the left from the object, fully characterize the final image in the system. In other words provide final image location, real/virtual, upright/inverted, final magnification, and final image height.
Explanation / Answer
case 1
first image will be made by lense
1/v -1/ u = 1/ f
1/v -1/ -20 = 1/ -20
v = -10 cm virtual erect
m1 = v/u = -10 / -20 = 0.5
distance from mirror = 45 -10 = 35 cm
1/v +1/u =1/ f
1/ v + 1/ -35 = 1/ -100
v = 53.84 cm virtual erect
m2 = -v/u = - 53.84 / -35 = 1.53 cm
final mag. M = m1*m2 = 0.5* 1.53 = 0.765
height of image = 0.765 * 5 = 3.82 cm high 53.84 right in the mirror virtual erect
case2
1/ v +1/ u = 1/f
u = 45-20 = 25
1/v + 1/ -25 = 1/ -100
v = 33.33 virtual erect
m1 = -v/u = - 33.33 / -25 = 1.33
distance from lense = 45 - 33.33 = 11.67 cm
1/v -1/u = 1/ f
1/ v -1/ -11.67 = 1/ -20
v = -7.36 cm virtual erect
m2 = v/u = -7.36 / -11.67 = 0.63
total mgnification = m1*m2 = 1.33* 0.63 = 0.837
height of image = 0.837*5 = 4.185 cm 7.36 cm left in lens virtual erect