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NEET PHYSICSEasy

An object is placed on the principal axis of a concave mirror at a distance of 1.5f1.5f (ff is the focal length). The image will be at:

A

-3f

B

1.5f

C

-1.5f

D

3f

Step-by-Step Solution

  1. Identify Formula: Use the mirror formula: 1v+1u=1f\frac{1}{v} + \frac{1}{u} = \frac{1}{f}.
  2. Sign Convention: According to the Cartesian sign convention for a concave mirror:
  • Object distance (uu) is taken as negative: u=−1.5fu = -1.5f.
  • Focal length (ff) is taken as negative for a concave mirror: Focal length = −f-f.
  1. Calculation: Substitute the values into the formula: 1v+1−1.5f=1−f\frac{1}{v} + \frac{1}{-1.5f} = \frac{1}{-f} 1v=11.5f−1f\frac{1}{v} = \frac{1}{1.5f} - \frac{1}{f} 1v=1015f−1f=23f−33f\frac{1}{v} = \frac{10}{15f} - \frac{1}{f} = \frac{2}{3f} - \frac{3}{3f} 1v=2−33f=−13f\frac{1}{v} = \frac{2 - 3}{3f} = \frac{-1}{3f} v=−3fv = -3f
  2. Conclusion: The image is formed at a distance of 3f3f from the mirror. The negative sign indicates that the image is formed on the same side as the object (real image).
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