A magician wishes to create the illusion of a 2.59-m-tall elephant. He plans to

Question

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

Explanation / Answer

(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)

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