A bi-convex lens of focal length is placed in air as shown in Figure 3(a) below. The radii of…
A bi-convex lens of focal length $f_{1}$ is placed in air as shown in Figure 3(a) below.
The radii of curvature of its first and second surfaces are R and 2R respectively. The lens is cut along the plane CD. Compare the focal length $f_{2}$ of the resulting lens shown in Figure 3(b) with that of the original lens shown in Figure 3(a).
Answer
Answer
AIWritten by AI - it can contain mistakes.
Using the Lens Maker's Formula:
$\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_{1}} - \frac{1}{R_{2}}\right)$
For the original bi-convex lens (Figure 3(a)):
$R_{1} = +R$ and $R_{2} = -2R$
$\frac{1}{f_{1}} = (\mu - 1)\left(\frac{1}{R} - \left(-\frac{1}{2R}\right)\right) = (\mu - 1)\frac{3}{2R}$
$f_{1} = \frac{2R}{3(\mu - 1)}$
For the cut lens (Figure 3(b)):
The first surface is plane ($R_{1}' = \infty$) and the second surface has radius $R_{2}' = -2R$:
$\frac{1}{f_{2}} = (\mu - 1)\left(\frac{1}{\infty} - \left(-\frac{1}{2R}\right)\right) = (\mu - 1)\frac{1}{2R}$
$f_{2} = \frac{2R}{\mu - 1}$
Comparing the focal lengths:
$\frac{f_{2}}{f_{1}} = \frac{\frac{2R}{\mu - 1}}{\frac{2R}{3(\mu - 1)}} = 3 \implies f_{2} = 3f_{1}$
Final answer: 3
Final answer: 3
From ISC 2026 Physics Paper 1, question 6.