Question #274549

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.


1
Expert's answer
2021-12-02T13:58:37-0500

The focal length of the mirror is

f=R2f = \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=24.02=12  cmf = \frac{24.0}{2} = 12 \; cm

The mirror equation is

1f=1d0+1di\frac{1}{f} = \frac{1}{d_0} +\frac{1}{d_i}

d0 = object distance

di = image distance

f = focal length of the mirror

1di=1f1do=doffdodi=fdodof=12×15.515.512=1863.5=53.14  cm\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=ddo=hihoM = \frac{-d}{d_o} = \frac{h_i}{h_o}

hi = height of the image

ho = height of the object

do = object distance

di = imege distance from the mirror

hi=h0×dido=0.3×53.1415.5=1.0285h_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.


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