In: Physics
A magician wishes to create the illusion of a 2.59-m-tall elephant. He plans to do this by forming a virtual image of a 74.0 cm tall model elephant with the help of a spherical mirror.
(a) Should the mirror be concave or convex?
concave / convex
Explain.
(b) If the model must be placed 3.00 m from the mirror, what radius
of curvature is needed? (Include the sign of each answer.)
____m
(c) How far from the mirror will the image be formed? (Enter a
negative value if the image forms behind the mirror.)
____m
(a) Should be concave mirror.
Toform a virtual, upright, and enlarged image, the mirror should be concave.
B_
When dealing with lenses and mirrors, there are two equationsthat are very useful:
M = -i /o and 1/f = 1/o + 1/i where
M is magnification, f is focal length, o is objectdistance, i is image distance
Notice that they give you that M is 2.59 / 0.74 =3.5 (i.e. the ratio of size of image tosize of object). It is + because the image is upright.
Also they give you that o = 3.00m So they give you M and o, you have tofind f and i.
Two equations, and two unknowns.
First, use M = - i /o or i = - Mo = - 3.5 * 3.00 m = -10.5 m (answer part c).
the negative image distance means that the image isbehind the mirror (a virtual image)
And now you can get f:
1/f = 1/o + 1/i or 1/f = (i + o ) / io flipover both sides
f = io / ( i + o) = -10.5* 3.00 / ( -10.5 + 3.00) = 4.2 m
The radius of curvcurva is twice the focal length so R = 2*4.2 = 8.4 meters (b - answer)