JEE Advanced · 2012 · 08 Apr 2012 · Paper 1 · Geometrical Optics

A bi-convex lens is formed with two thin plano-convex lenses as shown in the figure. Refractive index

n

of the first lens is 1.5 and that of the second lens is 1.2. Both the curved surfaces are of the same radius of curvature

R = 14 cm

. For this bi-convex lens, for an object distance of 40 cm, the image distance will be

(A)-280.0 cm.
(B)40.0 cm.✓ Correct
(C)21.5 cm.
(D)13.3 cm .

Answer: (B)

Solution

The total power is

P_{T} = (1.5 - 1) (\frac{1}{14}) + (1.2 - 1) (\frac{1}{14}) = \frac{0.5}{14} + \frac{0.2}{14} = \frac{0.7}{14} = \frac{1}{20}

. So

f = 20

cm. Using the lens formula:

\frac{1}{v} - \frac{1}{- 40} = \frac{1}{20}

, giving

\frac{1}{v} = \frac{1}{20} - \frac{1}{40} = \frac{1}{40}

, so

v = 40

cm.

JEE Advanced · 2012 · 08 Apr 2012 · Paper 1 · Single correct option

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