The object in the figure below is midway between the lens and the mirror, which
ID: 1499389 • Letter: T
Question
The object in the figure below is midway between the lens and the mirror, which are separated by a distance d = 24.8 cm. The magnitude of the mirror's radius of curvature is 19.8 cm, and the lens has a focal length of ?15.8 cm.
(a) Considering only the light that leaves the object and travels first toward the mirror, locate the final image formed by this system. .496 Incorrect: Your answer is incorrect. Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. cm to the right of the lens
(b) Is the image real or virtual? real virtual
(c) Is it upright or inverted? upright inverted
(d) What is the overall magnification of the image?
Explanation / Answer
a) For mirror:
1/do + 1/di1 = 1/fm
=> 1/12.4 + 1/di1 = 1/(19.8/2)
=> di1 = 9.9 * 12.4 / (12.4 - 9.9) = 49.1 cm
magnification by mirror, m1 = hi1/ho = -di1/do = -49.1/12.4 = -3.96
Negative magnification means the mirror creates an inverted image.
For lens:
Object distance = -(di1 - d)= -(49.1 - 24.8) cm = -24.3 cm
1/(-24.3) + 1/di = 1/(-15.8)
=> di = 15.8 * 24.3 / (15.8 - 24.3) = -45.2 cm
So, the final image is 45.2 cm to the right of the lens. (This is because for the lens, negative side is the right side since the incident ray is moving to the left). Negative sign of di means image is virtual.
magnification by lens, m2 = hi/hi1 = -(-45.2) / (-24.3) = -1.86
Negative magnification means the lens creates an inverted image.
b) Image is virtual.
c) Image is erect. (Mirror creates an inverted image and the lens inverts it again to form an erect image)
d) Overall magnification, m = m1m2 = (-3.96) * (-1.86) = 7.37