A convex lens forms a real image at $60$ cm on one side of an object placed $20$ cm in front of it. Find the focal length of the lens.
A$10$ cm
B$12$ cm
C$15$ cm
D$20$ cm
Answer & Solution
Correct answer: C. $15$ cm
Lens formula: $\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$. Sign convention: $u = -20$ cm (object on the left), $v = +60$ cm (real image on the right).
$\dfrac{1}{f} = \dfrac{1}{60} - \dfrac{1}{-20} = \dfrac{1}{60} + \dfrac{3}{60} = \dfrac{4}{60} = \dfrac{1}{15}$
So $f = 15$ cm.
**Trap.** Forgetting the sign on $u$ gives $\dfrac{1}{f} = \dfrac{1}{60} - \dfrac{1}{20} = -\dfrac{1}{30}$, leading to a negative $f$ — that's how option B (12) or A (10) creeps in.
Related questions
Two thin convex lenses, each of focal length 20 cm, are placed coaxially 10 cm apart in aiA Galilean telescope (objective convex, eyepiece concave) has objective focal length 20 cmA thin prism of refracting angle 5° is made of glass with n = 1.5. The deviation δ of a liRefraction at a single convex spherical surface: an object in air (n₁ = 1) at distance 20 A converging lens of focal length f is split along its principal axis into two semicirculaA compound microscope has an objective of focal length 1.0 cm and an eyepiece of focal lenGlass has refractive index 1.5 with respect to air. A light ray inside the glass strikes tThe apparent depth of an object lying at the bottom of a pool of water (n = 4/3) is, in te