Two convex lenses (L_1 and L_2) of equal foca | vedclass

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(D) For the first lens $L_1$, the object distance $u_1 = -4f$ and focal length $f_1 = f$. Using the lens formula $\frac{1}{v_1} - \frac{1}{u_1} = \fra

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$5f$ to the left of $L_2$

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(D) For the first lens $L_1$, the object distance $u_1 = -4f$ and focal length $f_1 = f$. Using the lens formula $\frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f_1}$:$\frac{1}{v_1} - \frac{1}{-4f} = \frac{1}{f} \Rightarrow \frac{1}{v_1} = \frac{1}{f} - \frac{1}{4f} = \frac{3}{4f} \Rightarrow v_1 = \frac{4f}{3}$.This image acts as an object for the second lens $L_2$. The distance between the lenses is $d = \frac{f}{2}$.The object distance for the second lens is $u_2 = v_1 - d = \frac{4f}{3} - \frac{f}{2} = \frac{8f - 3f}{6} = \frac{5f}{6}$.Since the object is to the left of $L_2$, $u_2 = -\frac{5f}{6}$.Using the lens formula for $L_2$ with $f_2 = f$:$\frac{1}{v_2} - \frac{1}{u_2} = \frac{1}{f_2} \Rightarrow \frac{1}{v_2} - \frac{1}{-5f/6} = \frac{1}{f} \Rightarrow \frac{1}{v_2} + \frac{6}{5f} = \frac{1}{f}$.$\frac{1}{v_2} = \frac{1}{f} - \frac{6}{5f} = \frac{5 - 6}{5f} = -\frac{1}{5f} \Rightarrow v_2 = -5f$.The negative sign indicates that the final image is formed at a distance $5f$ to the left of $L_2$.

Two convex lenses ($L_1$ and $L_2$) of equal focal length $f$ are placed at a distance $\frac{f}{2}$ apart. An object is placed at a distance $4f$ to the left of $L_1$ as shown in the figure. The final image is at:

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