A point object, 'O' is placed in front of two thin symmetrical coaxial convex lenses L₁ and L₂ with focal length 24 cm and 9 cm respectively. The distance between two lenses is 10 cm and the object is placed 6 cm away from lens L₁ as shown in the figure. The distance between the object and the image formed by the system of two lenses is __________ cm. Numerical answer type. Work it out, then open the answer.

Answer: 34. L₁: u = -6, f = 24: 1/v = 1/24 - 1/6 = -1/8 ⇒ v = -8 cm (virtual, 8 cm left of L₁). For L₂: u = -(8 + 10) = -18, f = 9: 1/v = 1/9 - 1/18 ⇒ v = 18 cm. Object to image = 6 + 10 + 18 = 34 cm

A point object, 'O' is placed in front of two thin symmetrical… | JEE Main 2023 (10 Apr, Shift 2)

Q37JEE Main 2023 · 10 APR · SHIFT 2Medium

Combination of Lenses

A point object, 'O' is placed in front of two thin symmetrical coaxial convex lenses $L_1$ and $L_2$ with focal length $24\text{ cm}$ and $9\text{ cm}$ respectively. The distance between two lenses is $10\text{ cm}$ and the object is placed $6\text{ cm}$ away from lens $L_1$ as shown in the figure. The distance between the object and the image formed by the system of two lenses is __________ $\text{cm}$ .

Numerical answer type. Work it out, then open the answer.

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Answer: 34

$L_1$ : $u = -6$ , $f = 24$ : $\frac{1}{v} = \frac{1}{24} - \frac{1}{6} = -\frac{1}{8} \Rightarrow v = -8\text{ cm}$ (virtual, 8 cm left of $L_1$ ).
For $L_2$ : $u = -(8 + 10) = -18$ , $f = 9$ : $\frac{1}{v} = \frac{1}{9} - \frac{1}{18} \Rightarrow v = 18\text{ cm}$ .
Object to image $= 6 + 10 + 18 = 34\text{ cm}$

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