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A convex mirror has an object distance of 2 m and a magnification of 1/3. Calculate the focal length of the mirror.

Physics

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Answer

1 m\text{1 m}

Step-by-step solution

3 steps
  1. Identify the given values and apply sign conventions.

    The object is placed in front of a convex mirror.
    Object distance, u=2u = 2 m. According to the Cartesian sign convention, for a real object, uu is negative.
    So, u=−2u = -2 m.
    Magnification, m=13m = \frac{1}{3}. For a convex mirror, the image is always virtual and erect, so the magnification is positive.
    We need to find the focal length, ff. For a convex mirror, the focal length ff is positive.

  2. Calculate the image distance (vv) using the magnification formula.

    The magnification formula for mirrors is:

    m=−vum = -\frac{v}{u}

    Substitute the known values:

    13=−v−2\frac{1}{3} = -\frac{v}{-2} 13=v2\frac{1}{3} = \frac{v}{2}

    Multiply both sides by 2 to solve for vv:

    v=23 mv = \frac{2}{3} \text{ m}

    The positive value of vv indicates that the image is virtual and formed behind the mirror, which is consistent with a convex mirror.

  3. Calculate the focal length (ff) using the mirror formula.

    The mirror formula is:

    1f=1u+1v\frac{1}{f} = \frac{1}{u} + \frac{1}{v}

    Substitute the values of uu and vv:

    1f=1−2+123\frac{1}{f} = \frac{1}{-2} + \frac{1}{\frac{2}{3}} 1f=−12+32\frac{1}{f} = -\frac{1}{2} + \frac{3}{2}

    Combine the fractions:

    1f=−1+32\frac{1}{f} = \frac{-1 + 3}{2} 1f=22\frac{1}{f} = \frac{2}{2} 1f=1\frac{1}{f} = 1 f=1 mf = 1 \text{ m}

    The positive value of ff is consistent with the focal length of a convex mirror.

    The focal length of the mirror is 1 m\text{1 m}.

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