JEE Main · 2024 · 08 Apr 2024 · Shift 2 · Geometrical Optics

The position of the image formed by the combination of lenses is :

(A)30 cm (right of third lens)✓ Correct
(B)15 cm (left of second lens)
(C)30 cm (left of third lens)
(D)15 cm (right of second lens)

Answer: (A)

Solution

Lens 1 (

f_{1} = 10

cm): object at

u = - 30

cm.

\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{10} - \frac{1}{30} = \frac{1}{15} ⟹ v = 15 cm

This image lies 15 cm to the right of lens 1, i.e.

15 - 5 = 10

cm beyond lens 2 — a virtual object for lens 2 at

u = + 10

cm.

Lens 2 (

f_{2} = - 10

cm):

\frac{1}{v} = \frac{1}{- 10} + \frac{1}{10} = 0 ⟹ v = \infty

Rays leave lens 2 parallel to the axis.

Lens 3 (

f_{3} = 30

cm) receives a parallel beam, so the final image forms at its focus, 30 cm to the right of lens 3.

JEE Main · 2024 · 08 Apr 2024 · Shift 2 · Single correct option

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