A double concave lens is made of glass of refractive index 1.5. Its focal length is 20 cm in magnitude. One face has twice the radius of curvature of the other face. The smaller of the two radii of curvature is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 15 cm
For a concave lens, \(R_1\) is negative and \(R_2\) is positive, so \(f\) comes out negative. Let the smaller radius be \(R\). Then the other face has radius \(2R\).
\( f=-20\ \mathrm{cm} \)
\( R_1=-R \)
\( R_2=+2R \)
\( \begin{aligned} \frac{1}{f} &= (n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \\ &= 0.5\left(-\frac{1}{R}-\frac{1}{2R}\right) \\ &= -\frac{3}{4R} \end{aligned} \)
\( \begin{aligned} -\frac{1}{20} &= -\frac{3}{4R} \\ R &= 15\ \mathrm{cm} \end{aligned} \)
The other face has a radius of 30 cm. Hence, the smaller radius of curvature is 15 cm.
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. It is useful to design lenses of desired focal length using surfaces of suitable radii of curvature. Note that the formula is true for a concave lens also. In that case R1 is negative, R2 positive and therefore, f is negative.”