Q.11.An object placed 10.2 cm from a concave spherical mirror produces a real im
ID: 2096863 • Letter: Q
Question
Q.11.An object placed 10.2 cm from a concave spherical mirror produces a real image 6.93 cmfrom the mirror. If the object is moved to a new position 20.4 cm from the mirror, what is the position of the image? Answer in units of cm.
Q.12.A spherical Christmas tree ornament is 4 cm in diameter. What is the magni?cation of the image of an object placed 10.9 cm away from the ornament?
Q.13.A convex mirror has a focal length of 22.4 cm. Find the object location for which the image will be one-half as tall as the object. Answer in units of cm.
Explanation / Answer
Q11: You basically have to use the same focal point from the previous position to find your new position.
Old position: (1/do) + (1/di) = (1/f) , where dois distance of object, di is distance of image, and f is your focal point.
(1/10.2) + (1/6.93) = (1/focal) = 0.242
New position: (1/20.4) + (1/image) = 0.242
New image position should be 5.182 cm
Q12: Magnification for convex mirrors is basically the negative of the distance of image over distance of object, (-di/do). But you don't have di so you must use (1/do) + (1/di) = (1/f)
Focal point is basically radius/2, so f = (4/2)/2 = 1, you assume it's a convex reflector, so the focal point is -1
(1/do) + (1/10.9) = (1/-1), where the image's position is determined to be -0.916 cm
then simply plug it into the magnification equation, m = (-0.916/-10.9) = 0.084
Q13:
Seeing as how the image is one-half as tall as the object, the magnification equation for this convex mirror looks like this:
m = -di/do = 1/2.
Thus, do = -2di.
Using the lens-mirror equation:
1/f = 1/di+ 1/do
(1/-22.4) = 1/di - 1/(2di)
Focal point MUST BE negative due to it being a convex mirror.
(1/-22.4) = 1/(2di)
di = -11.2 cm
Therefore, do = 22.4 cm
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