An object is located at a distance of 115 cm from a concave mirror of focal leng
ID: 1377758 • Letter: A
Question
An object is located at a distance of 115 cm from a concave mirror of focal length 21.5 cm. Another concave mirror of focal length 6.81 cm is located 21.5 cm in front of the first concave mirror. The reflecting sides of the two mirrors face each other. Assume the second mirror is partially silvered so that the light rays from the object go through it, reflect from the first mirror, and then reflect from the second mirror.
a) What is the location of the final image formed by the two mirrors? Give your answer in terms of the distance from the second mirror.
b) What is the total magnification produced by the combination of the two mirrors?
Explanation / Answer
Given that,
focal length of first concave mirror = f1 = 21.5 cm ; object distance from the first mirror = o1 = 115 cm
Focal length of the second mirror = f2 = 6.81 cm and distance between the mirrors = d = 21.5 cm
(a)
The lens eqn for the first mirror will be as follos:
1/f1 = 1/i1 + 1/o1
i1 = o1 x f1 / (o1 - f1) = 115 x 21.5 / (115 - 21.5) = 2472.5 / 93.5 = 26.44 cm
hence i1 = 26.44 cm
The image generated by the first concave mirror will act as the object for the second one, so we have
o2 = i1 - d = 21.5 - 26.44 = - 4.94 cm
f2 = 6.81 cm
so for second mirror we can write,
1/f2 = 1/i2 + 1/o2
i2 = o2 x f2 / (o2 - f2) = -4.94 x 6.81 / -4.94 -6.81 = - 33.64 / -11.75 = 2.86 cm
i2 = 2.86 cm cm behind the second mirror
(b)magnification is given by
m = -i/o
m1 = -i1/o1 = 26.44/115 = 0.23
m2 = -i2/o2 = -2.86/4.94 = 0.58
magnitfication in terms of height
In this case,
hi1 =ho2
hi2 = m2 x hi1
hi2 = m2 x m1 x ho1
hi2 = (-0.58) ( -0.23) ho1
hi2 =+ 0.13(ho1)
hence the total magnification produced by the combination of two mirros is 0.84 and its a positive number. which means that the final image is not inverted.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.