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3 parts
a

16.67 cm16.67 \text{ cm}

b

∞\infty

c

−10 cm-10 \text{ cm}

Step-by-step solution

9 steps

The problem involves a concave spherical mirror with a focal length f=+10 cmf = +10 \text{ cm}. We will use the mirror formula and magnification formula to calculate the image distance and describe the image for different object distances.

The mirror formula is:

1f=1do+1di\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}

where ff is the focal length, dod_o is the object distance, and did_i is the image distance.
Rearranging for did_i:

1di=1f−1do\frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} di=f⋅dodo−fd_i = \frac{f \cdot d_o}{d_o - f}

The magnification MM is given by:

M=−didoM = -\frac{d_i}{d_o}
  • If di>0d_i > 0, the image is real. If di<0d_i < 0, the image is virtual.
  • If M<0M < 0, the image is inverted. If M>0M > 0, the image is upright.
  • If ∣M∣>1|M| > 1, the image is magnified. If ∣M∣<1|M| < 1, the image is diminished. If ∣M∣=1|M| = 1, the image is the same size.

Given: f=+10 cmf = +10 \text{ cm}

Part a· 3 steps

Object distance do=25 cmd_o = 25 \text{ cm}

  1. Calculate the image distance did_i.

    1di=110 cm−125 cm\frac{1}{d_i} = \frac{1}{10 \text{ cm}} - \frac{1}{25 \text{ cm}}

    Find a common denominator (50):

    1di=550 cm−250 cm\frac{1}{d_i} = \frac{5}{50 \text{ cm}} - \frac{2}{50 \text{ cm}} 1di=350 cm\frac{1}{d_i} = \frac{3}{50 \text{ cm}} di=503 cmd_i = \frac{50}{3} \text{ cm} di≈16.67 cmd_i \approx 16.67 \text{ cm}

  2. Calculate the magnification MM.

    M=−dido=−50/3 cm25 cmM = -\frac{d_i}{d_o} = -\frac{50/3 \text{ cm}}{25 \text{ cm}} M=−503×25=−23M = -\frac{50}{3 \times 25} = -\frac{2}{3}

  3. Describe the image.

    Since did_i is positive, the image is real.
    Since MM is negative, the image is inverted.
    Since ∣M∣=2/3<1|M| = 2/3 < 1, the image is diminished.

    The image distance is 16.67 cm\boxed{16.67 \text{ cm}}. The image is real, inverted, and diminished.

Part b

∞\infty

Object distance do=10 cmd_o = 10 \text{ cm}

  1. Calculate the image distance did_i.

    1di=110 cm−110 cm\frac{1}{d_i} = \frac{1}{10 \text{ cm}} - \frac{1}{10 \text{ cm}} 1di=0\frac{1}{d_i} = 0

    This implies di→∞d_i \rightarrow \infty.

  2. Calculate the magnification MM.

    M=−dido=−∞10 cm=−∞M = -\frac{d_i}{d_o} = -\frac{\infty}{10 \text{ cm}} = -\infty

  3. Describe the image.

    When the object is placed at the focal point of a concave mirror, the reflected rays are parallel and form an image at infinity.
    The image is real, inverted, and highly magnified (at infinity).

    The image distance is ∞\boxed{\infty}. The image is real, inverted, and highly magnified.

Part c

−10 cm-10 \text{ cm}

Object distance do=5 cmd_o = 5 \text{ cm}

  1. Calculate the image distance did_i.

    1di=110 cm−15 cm\frac{1}{d_i} = \frac{1}{10 \text{ cm}} - \frac{1}{5 \text{ cm}}

    Find a common denominator (10):

    1di=110 cm−210 cm\frac{1}{d_i} = \frac{1}{10 \text{ cm}} - \frac{2}{10 \text{ cm}} 1di=−110 cm\frac{1}{d_i} = -\frac{1}{10 \text{ cm}} di=−10 cmd_i = -10 \text{ cm}

  2. Calculate the magnification MM.

    M=−dido=−−10 cm5 cmM = -\frac{d_i}{d_o} = -\frac{-10 \text{ cm}}{5 \text{ cm}} M=+2M = +2

  3. Describe the image.

    Since did_i is negative, the image is virtual.
    Since MM is positive, the image is upright.
    Since ∣M∣=2>1|M| = 2 > 1, the image is magnified.

    The image distance is −10 cm\boxed{-10 \text{ cm}}. The image is virtual, upright, and magnified.

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