Answer to Question #274549 in Optics for Chest

Question

1.    An object 0.3 cm tall is placed 15.5 cm to the left of the vertex of a concave mirror with a center of curvature of 24.0 cm. Determine the position, size, orientation and type (real or virtual) of the image.

Answer 1

The focal length of the mirror is

$f = \frac{R}{2}$

R = radius of curvature of the mirror

For the convace mirror the radius of curvature of the mirror will be positive:

$f = \frac{24.0}{2} = 12 \; cm$

The mirror equation is

$\frac{1}{f} = \frac{1}{d_0} +\frac{1}{d_i}$

d₀ = object distance

d_i = image distance

f = focal length of the mirror

$\frac{1}{d_i} = \frac{1}{f} -\frac{1}{d_o} = \frac{d_o-f}{fd_o} \\ d_i = \frac{fd_o}{d_o-f} \\ = \frac{12 \times 15.5}{15.5 -12} = \frac{186}{3.5} = 53.14 \;cm$

The imagei s formed at a distance of 53.14 cm to the left of mirror vertex.

The magnification produced by the mirror is

$M = \frac{-d}{d_o} = \frac{h_i}{h_o}$

h_i = height of the image

h_o = height of the object

d_o = object distance

d_i = imege distance from the mirror

$h_i = h_0 \times \frac{-d_i}{d_o} \\ = 0.3 \times \frac{-53.14}{15.5} = -1.0285$

As the height of the image is negative the image formed will be inverted.

As the distance is positive the image will be real.