An object is placed 11 cm in front of a concave mirror whose focal length is 23 cm. The object is 3.6 cm tall. Determine (a) the location of the image, taking a real image as a positive value and a virtual image as a negative value. (b) Determine the height of the image, where an upright image will have a positive height and an inverted image will have a negative height.
To determine the location and height of the image formed by a concave mirror, we can use the mirror equation and magnification formula.
(a) Location of the image:
The mirror equation relates the object distance (denoted as "do"), the image distance (denoted as "di"), and the focal length (denoted as "f") of the mirror:
1/do + 1/di = 1/f
Given:
do = -11 cm (since the object is placed in front of the mirror)
f = -23 cm (since the focal length of a concave mirror is negative)
Substituting these values into the mirror equation:
1/-11 + 1/di = 1/-23
Simplifying the equation gives:
1/di = 1/-23 + 1/11 = -34/253
To find the value of di, take the reciprocal of both sides:
di = -253/34 ≈ -7.441 cm
Since the equation yields a negative value, the image formed by the concave mirror is virtual.
Therefore, the location of the image is approximately -7.441 cm behind the mirror, which means it is formed in front of the mirror.
(b) Height of the image:
The magnification formula relates the height of the image (denoted as "hi") and the height of the object (denoted as "ho") to the magnification (denoted as "m"):
m = -hi/ho = di/do
Given:
ho = 3.6 cm
Substituting the calculated values:
m = (-7.441)/(-11) ≈ 0.6765
Now, using the magnification formula, solve for hi:
hi = m * ho ≈ 0.6765 * 3.6 ≈ 2.435 cm
The calculated height of the image is approximately 2.435 cm.
Therefore:
(a) The location of the image is approximately -7.441 cm (virtual image formed in front of the mirror).
(b) The height of the image is approximately 2.435 cm (upright image).