The position of the final image formed by the | vedclass

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(D) For the first lens:$f_1 = +10 \ cm, u_1 = -30 \ cm$.Applying the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$:$\frac{1}{v_1} = \frac{1}{

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$30 \ cm$

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(D) For the first lens:$f_1 = +10 \ cm, u_1 = -30 \ cm$.Applying the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$:$\frac{1}{v_1} = \frac{1}{f_1} + \frac{1}{u_1} = \frac{1}{10} - \frac{1}{30} = \frac{3-1}{30} = \frac{2}{30} = \frac{1}{15}$.So, $v_1 = +15 \ cm$.This image acts as an object for the second lens. The distance between the first and second lens is $5 \ cm$, so the object distance for the second lens is $u_2 = +(15 - 5) = +10 \ cm$.For the second lens $(f_2 = -10 \ cm)$:$\frac{1}{v_2} = \frac{1}{f_2} + \frac{1}{u_2} = \frac{1}{-10} + \frac{1}{10} = 0$.So, $v_2 = \infty$.The rays emerging from the second lens are parallel to the principal axis.These parallel rays fall on the third lens $(f_3 = +30 \ cm)$. Parallel rays incident on a convex lens converge at its focus.Therefore, the final image is formed at a distance of $30 \ cm$ from the third lens.

The position of the final image formed by the given lens combination from the third lens will be at a distance of $(f_1 = +10 \ cm, f_2 = -10 \ cm, f_3 = +30 \ cm)$.

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