An object is placed in front of an equiconvex lens of focal length \( 10\,\mathrm{cm} \). The two object distances for which image will be four times the size of object are:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( -7.5\,\mathrm{cm} \) and \( -12.5\,\mathrm{cm} \)
Use the lens formula \( 1/v - 1/u = 1/f \) with the Cartesian sign convention (\( f = +10\,\mathrm{cm} \) for a convex lens, object distances negative) and \( m = v/u \). An image four times the size of the object can be real and inverted (\( m = -4 \)) or virtual and erect (\( m = +4 \)).
Putting \( v = mu \):
\( \displaystyle \frac{1}{mu} - \frac{1}{u} = \frac{1}{f} \Rightarrow u = \frac{f(1-m)}{m} \)
\( \displaystyle m = -4:\quad u = \frac{10\,\mathrm{cm}\times(1+4)}{-4} = -12.5\,\mathrm{cm} \)
\( \displaystyle m = +4:\quad u = \frac{10\,\mathrm{cm}\times(1-4)}{4} = -7.5\,\mathrm{cm} \)
The object is at 12.5 cm (between \( f \) and \( 2f \)) for the real image and at 7.5 cm (inside \( f \)) for the virtual image.