Question

find the position of the image formed by the lens combination ?

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Answer 1

Answer:

please send the picture properly

Explanation:

Answer 2

Answer:

The image will be formed 30cm to the right of the third lens

Explanation:

• Apply Lens Formula :-

$\frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f_1}$

Image formed by the first lens is

Putting in the values

$\implies \frac{1}{v_1} + \frac{1}{ - 30} = \frac{1}{10} \\ \\ \implies \frac{1}{v_1} = \frac{1}{10} + \frac{1}{30} \\ \\ \implies \frac{1}{v_1} = \frac{1}{15} \\ \\ \implies v_1 = 15 \: cm$

Now , this image formed will act as the object for the second lens.

So, Object distance will be 15 - 5 = 10 cm .

Conclusion :-

Since this image is real then it will act as virtual object for the Second lens that means the rays do not actually meet but appers to come from it for second lens.

Now, again using the lens Formula,

$\implies \frac{1}{v_2} - \frac{1}{10} = \frac{1}{ - 10} \\ \\ \implies \frac{1}{v_2} = \frac{1}{ - 10} + \frac{1}{10} \\ \\ \implies v_2 = ∞$

Conclusion:-

The virtual image is formed which is at an infinite distance to the left side of the second lens and now this will act as an object for the third lens.

Now , By again using the lens formula

$\implies \frac{1}{v_3} - \frac{1}{u_3} = \frac{1}{f_3} \\ \\ \implies \frac{1}{v_3} = \frac{1}{f_3} + \frac{1}{u_3} \\ \\ \implies \frac{1}{v_3} = \frac{1}{ \infty } + \frac{1}{30} \\ \\ \implies \frac{1}{v_3} = \frac{1}{30} \\ \\ \implies V_3 = 30 \: cm$

here, we get $V_3$ as 30 cm that is final image distance .