A point object,O is placed in front of two th | vedclass

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(C) For the first lens $L_1$,the object distance $u_1 = -6\,cm$ and focal length $f_1 = +24\,cm$.Using the lens formula $\frac{1}{v_1} - \frac{1}{u_1}

Vedclass · Accepted Answer

$34$

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(C) For the first lens $L_1$,the object distance $u_1 = -6\,cm$ and focal length $f_1 = +24\,cm$.Using the lens formula $\frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f_1}$:$\frac{1}{v_1} - \frac{1}{-6} = \frac{1}{24}$$\frac{1}{v_1} = \frac{1}{24} - \frac{1}{6} = \frac{1-4}{24} = -\frac{3}{24} = -\frac{1}{8}$So,$v_1 = -8\,cm$. This means a virtual image $I_1$ is formed $8\,cm$ to the left of $L_1$.For the second lens $L_2$,the object distance $u_2$ is the distance of $I_1$ from $L_2$. Since $I_1$ is $8\,cm$ to the left of $L_1$ and $L_1$ is $10\,cm$ to the left of $L_2$,the distance $u_2 = -(8 + 10) = -18\,cm$. The focal length $f_2 = +9\,cm$.Using the lens formula $\frac{1}{v_2} - \frac{1}{u_2} = \frac{1}{f_2}$:$\frac{1}{v_2} - \frac{1}{-18} = \frac{1}{9}$$\frac{1}{v_2} = \frac{1}{9} - \frac{1}{18} = \frac{2-1}{18} = \frac{1}{18}$So,$v_2 = +18\,cm$. This means the final image is formed $18\,cm$ to the right of $L_2$.The object is $6\,cm$ to the left of $L_1$. The final image is $18\,cm$ to the right of $L_2$. The distance between the lenses is $10\,cm$.The total distance between the object and the final image is $6\,cm + 10\,cm + 18\,cm = 34\,cm$.

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