Find the position of the final image formed b | vedclass

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(A) For the first lens $(f_1 = 10 \ cm)$: The object distance is $u_1 = -30 \ cm$. Using the lens formula $\frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f_

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$30 \ cm$ (right of third lens)

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(A) For the first lens $(f_1 = 10 \ cm)$: The object distance is $u_1 = -30 \ cm$. Using the lens formula $\frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f_1}$,we get $\frac{1}{v_1} = \frac{1}{10} - \frac{1}{30} = \frac{3-1}{30} = \frac{2}{30} = \frac{1}{15}$. Thus,$v_1 = 15 \ cm$. The image is formed $15 \ cm$ to the right of the first lens.For the second lens $(f_2 = -10 \ cm)$: The distance between the first and second lens is $5 \ cm$. The image formed by the first lens acts as an object for the second lens. The object distance is $u_2 = +(15 - 5) = +10 \ cm$. Using the lens formula $\frac{1}{v_2} - \frac{1}{u_2} = \frac{1}{f_2}$,we get $\frac{1}{v_2} = \frac{1}{-10} + \frac{1}{10} = 0$. Thus,$v_2 = \infty$.For the third lens $(f_3 = 30 \ cm)$: The rays are parallel to the principal axis as they come from infinity. Therefore,the final image is formed at the focal point of the third lens,which is $30 \ cm$ to the right of the third lens.

Find the position of the final image formed by the combination of the three lenses shown in the figure. The focal lengths are $f_1 = 10 \ cm$,$f_2 = -10 \ cm$,and $f_3 = 30 \ cm$.

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