A thin glass lens (refractive index 1.5) has radii of curvature \(R_1=+10\ \mathrm{cm}\) and \(R_2=+20\ \mathrm{cm}\), as per the Cartesian sign convention. The power of the lens is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → +2.5 D
The lens maker’s formula gives the focal length from the two radii, with their signs. Here both radii are positive, because both centres of curvature lie on the same side of the lens. So the two terms subtract. The power is \(P=1/f\), with \(f\) in metres.
\( n=1.5 \)
\( R_1=+10\ \mathrm{cm} \)
\( R_2=+20\ \mathrm{cm} \)
\( \begin{aligned} \frac{1}{f} &= (n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \\ &= 0.5\left(\frac{1}{10}-\frac{1}{20}\right) \\ &= \frac{1}{40}\ \mathrm{cm^{-1}} \end{aligned} \)
\( \begin{aligned} f &= +40\ \mathrm{cm} \\ &= +0.4\ \mathrm{m} \end{aligned} \)
\( \begin{aligned} P &= \frac{1}{f} \\ &= \frac{1}{0.4} \\ &= +2.5\ \mathrm{D} \end{aligned} \)
The positive sign shows a converging lens. Hence, the power of the lens is +2.5 D.
NCERT (2026-27):
Refraction by a lens – “… By the sign convention, BC1 = + R1, DC2 = –R2 So … can be written as”
\( \displaystyle \frac{1}{f}=(n_{21}-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \)\( \qquad (9.21) \)
“Equation (9.21) is known as the lens maker’s formula. …”
Power of a lens – “… Thus,”
\( \displaystyle P=\frac{1}{f} \)\( \qquad (9.25) \)
“The SI unit for power of a lens is dioptre (D): 1D = 1m–1. …”