A thin converging lens L1 forms a real image of an object located far away from the lens as shown in the figure. The image is located a distance 4l and has height h. A diverging lens of focal length l is placed 2l from lens L1. Another converging lens of focal length 2l is placed 3l from lens L1. The height of final image thus formed is (Both diverging and converging lenses are placed at right side of L1 -

From 2nd lens 1v−12ℓ=1−ℓ or v=−2ℓm1 = -1From 3rd lens 1v−1−3ℓ=12ℓ or v=6ℓm2 = -2hI = (m1 × m2) h0= 2h

A thin converging lens L1 forms a real image of an object located far away from the lens as shown in the figure. The image is located a distance 4l and has height h. A diverging lens of focal length l is placed 2l from lens L1. Another converging lens of focal length 2l is placed 3l from lens L1. The height of final image thus formed is (Both diverging and converging lenses are placed at right side of L1 -

From 2nd lens 1v−12ℓ=1−ℓ or v=−2ℓm1 = -1From 3rd lens 1v−1−3ℓ=12ℓ or v=6ℓm2 = -2hI = (m1 × m2) h0= 2h

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A thin converging lens L₁ forms a real image of an object located far away from the lens as shown in the figure. The image is located a distance 4l and has height h. A diverging lens of focal length l is placed 2l from lens L₁. Another converging lens of focal length 2l is placed 3l from lens L₁. The height of final image thus formed is (Both diverging and converging lenses are placed at right side of L₁ -

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$\frac{h}{2}$

answer is D.

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Detailed Solution

From 2^nd lens $\frac{1}{v} - \frac{1}{2 ℓ} = \frac{1}{- ℓ} or v = - 2 ℓ$
m₁ = -1
From 3rd lens $\frac{1}{v} - \frac{1}{- 3 ℓ} = \frac{1}{2 ℓ} or v = 6 ℓ$
m₂ = -2
h_I= (m₁ × m₂) h₀
= 2h

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