A double convex lens made of glass of refractive index \( \frac{3}{2} \) has both radii of curvature of magnitude 20 cm. An object 3 cm high is placed at 10 cm from the lens. What is the nature and size of the image?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 6 cm; virtual and erect
The focal length follows from the lens maker’s formula, and the image from the lens formula with the Cartesian sign convention (distances measured from the lens, positive along the incident light). For a double convex lens \( R_1 = +20\,\mathrm{cm} \) and \( R_2 = -20\,\mathrm{cm} \).
\( \displaystyle \frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = \left(\frac{3}{2} - 1\right)\left(\frac{1}{20} + \frac{1}{20}\right) = \frac{1}{20}\,\mathrm{cm^{-1}} \)
So \( f = +20\,\mathrm{cm} \). The object is on the incident side, \( u = -10\,\mathrm{cm} \).
\( \displaystyle \frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{20} - \frac{1}{10} = -\frac{1}{20} \Rightarrow v = -20\,\mathrm{cm} \)
A negative \( v \) means the image is on the same side as the object, so it is virtual.
\( \displaystyle m = \frac{v}{u} = \frac{-20}{-10} = +2 \)
The positive magnification means an erect image, and its height is \( h' = m\,h = 2\times3\,\mathrm{cm} = 6\,\mathrm{cm} \).
Shortcut: For \( \mu = 1.5 \) a symmetric double convex lens has \( f = R \), so here \( f = 20\,\mathrm{cm} \). An object placed inside the focal length (10 cm < 20 cm) always gives a virtual, erect, magnified image, which leaves only Option (2) among the choices; at \( u = f/2 \) the image distance is \( v = -f \) and \( m = 2 \).