Two identical symmetric double convex lenses of focal length $f$ are cut into two equal parts $\mathrm{L}_{1}, \mathrm{~L}_{2}$ by AB plane and $\mathrm{L}_{3}, \mathrm{~L}_{4}$ by XY plane as shown in figure respectively. The ratio of focal lengths of lenses $\mathrm{L}_{1}$ and $\mathrm{L}_{3}$ is

JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Geometrical Optics

Two identical symmetric double convex lenses of focal length

f

are cut into two equal parts

L_{1}, L_{2}

by AB plane and

L_{3}, L_{4}

by XY plane as shown in figure respectively. The ratio of focal lengths of lenses

L_{1}

and

L_{3}

Structure Lens cutting diagrams — Lens cutting diagrams

Structure Alternative representation of lens cuts — Alternative representation of lens cuts

(A) $1 : 4$
(B) $1 : 1$
(C) $2 : 1$
(D) $1 : 2$ ✓ Correct

Answer: (D)

Solution

Cutting the lens along the AB plane (a plane containing the optical axis) splits it into two halves side by side,

L_{1}

and

L_{2}

; each half keeps both of its curved surfaces with their original radii of curvature, only the aperture (diameter) is reduced. Since the lens maker's formula

\frac{1}{f} = (μ - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})

does not involve the aperture, the focal length of each half is unchanged:

f_{L_{1}} = f_{L_{2}} = f

Cutting along the XY plane (perpendicular to the axis) splits the lens into a front piece and a back piece,

L_{3}

and

L_{4}

; each is now a plano-convex lens, keeping only one of the two original curved surfaces (radius

R

) with the other face flat. For the original symmetric biconvex lens,

R_{1} = R

R_{2} = - R

, so

\frac{1}{f} = (μ - 1) (\frac{1}{R} + \frac{1}{R}) = \frac{2 ( μ - 1 )}{R}

For the plano-convex half, only one curved surface contributes:

\frac{1}{f _{L_{3}}} = (μ - 1) \frac{1}{R} = \frac{1}{2 f} ⟹ f_{L_{3}} = 2 f

\frac{f _{L_{1}}}{f _{L_{3}}} = \frac{f}{2 f} = \frac{1}{2}

JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Single correct option

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JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Geometrical Optics

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