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

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Q.
The position of the image formed by the combination of lenses is [2024]

1 30 cm (left of third lens)

2 15 cm (right of second lens)

3 15 cm (left of second lens)

4 30 cm (right of third lens)

Ans.
(4)

Here we will use the lens formula three times, for lens 1

$u = - 30 cm, f = 10 cm$

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{v_{1}} - \frac{1}{- 30} = \frac{1}{10} \Rightarrow v_{1} = 15 cm$

For 2nd lens $u = 10 cm, f = - 10 cm$

$v_{2} \to \infty$

For 3rd lens: $f = 30 cm, u = - \infty, v = ?$

So $v$ will be 30 cm.

Solution

Q.
The position of the image formed by the combination of lenses is [2024]

1 30 cm (left of third lens)

2 15 cm (right of second lens)

3 15 cm (left of second lens)

4 30 cm (right of third lens)

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Solution

Q. The position of the image formed by the combination of lenses is [2024]

1 30 cm (left of third lens)

2 15 cm (right of second lens)

3 15 cm (left of second lens)

4 30 cm (right of third lens)

Ans. (4) Here we will use the lens formula three times, for lens 1 u=-30 cm, f=10 cm 1v-1u=1f⇒1v1-1-30=110⇒v1=15 cm For 2nd lens u=10 cm, f=-10 cm v2→∞ For 3rd lens: f=30 cm, u=-∞, v=? So v will be 30 cm.

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Q.
The position of the image formed by the combination of lenses is [2024]