Question:medium

A concave mirror has a focal length of \( 10 \, \text{cm} \). An object is placed at a distance of \( 15 \, \text{cm} \) from the mirror. Calculate the position of the image formed.

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For a concave mirror, if the object is placed outside the focal point, the image formed will be real and inverted. If placed inside the focal point, the image will be virtual and erect.
Updated On: Jan 13, 2026
  • \( 30 \, \text{cm} \) (real and inverted)
  • \( 5 \, \text{cm} \) (virtual and erect)
  • \( 10 \, \text{cm} \) (real and inverted)
  • \( 20 \, \text{cm} \) (virtual and erect)
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The Correct Option is A

Solution and Explanation

The mirror equation, which establishes a relationship between object distance \( u \), image distance \( v \), and focal length \( f \) of a mirror, is given by: \[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \] The following values are provided: - \( f = -10 \, \text{cm} \) (focal length of the concave mirror; negative as it is concave), - \( u = -15 \, \text{cm} \) (object distance; negative as the object is in front of the mirror), - \( v \) represents the image distance, which needs to be calculated. To find \( v \), the mirror equation is rearranged as follows: \[ \frac{1}{v} = \frac{1}{f} - \frac{1}{u} \] Substituting the known values yields: \[ \frac{1}{v} = \frac{1}{-10} - \frac{1}{-15} \] \[ \frac{1}{v} = -\frac{1}{10} + \frac{1}{15} = -\frac{3}{30} + \frac{2}{30} = -\frac{1}{30} \] Consequently, the image distance is: \[ v = -30 \, \text{cm} \] The negative sign for \( v \) indicates that the image is formed on the same side as the object, signifying a real and inverted image. The image is located \( 30 \, \text{cm} \) from the mirror.
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