Question

A). Find the two locations where an object can be placed in front of a concave mirror with a radius of curvature of 36cm such that its image is twice its size. answer in cm

B). In the case when the object is placed closer to the mirror, state whether the image is real or virtual, upright or inverted.

C). In the case when the object is placed farther to the mirror, state whether the image is real or virtual, upright or inverted.

Answer 1

A) Use the mirror equation and the magnification equation where m is the magnification, s is the distance that the object is from the mirror, is the image distance from the mirror, R is the radius of curvature, y is the height of object and is the height of the image.

Step 1) Use the magnification equation to get an expression for in terms of s. Since , .

Step 2) Plug in into and solve for the object distance s.

Step 3) The magnification m can be two different values in order for the image to be twice as tall is the object. Those values are . Focus on the case. Plug and into to find the object distance.

The object should be placed 9 cm in front of the mirror.

Step 4) Now repeat step 3 but instead using .

The object could also be placed 27 cm in front of the mirror.

B)

Step 5) In the case that the object is placed at (closer to the mirror), which was when the magnification was a positive 2, so . Since y is positive, must also be positive. A positive height means that the image is upright.

Step 6) Also from the magnification equation, so . Since s is positive, must be negative. This means the image is behind the mirror, so the image is virtual.

C)

Step 7) In the case that the object is placed at (further from the mirror), which was when the magnification was a negative 2, so . Since y is positive, must be negative. A negative height means that the image is inverted.

Step 8) Also from the magnification equation, so . Since s is positive, must also be positive. This means the image is in front of the mirror, so the image is real.