At what distance should an object be placed from a convex lens of foc…

The correct answer is Option (2) → 20 cm

A convex lens forms a real image that is inverted, so the magnification is negative: \( m = v/u = -3 \). Using the Cartesian sign convention (distances measured from the lens, positive in the direction of the incident light), the object distance \( u \) is negative.

    \( u = -x \)
    \( v = 3x \)
    \( f = +15\,\mathrm{cm} \)

By the lens formula,

    \( \displaystyle \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \)

    \( \displaystyle \frac{1}{3x} + \frac{1}{x} = \frac{1}{15} \;\Rightarrow\; \frac{4}{3x} = \frac{1}{15} \)

    \( \displaystyle x = \frac{4 \times 15}{3} = 20\,\mathrm{cm} \)

So the object is placed 20 cm from the lens, between \( f \) and \( 2f \), as a magnified real image requires.