Two lenses A and B having focal lengths 2.0 , | vedclass

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(B) Given: Focal length of lens $A$, $f_1 = 2 \,cm$. Focal length of lens $B$, $f_2 = 5 \,cm$. Distance between lenses $d = 14 \,cm$. Object distance

Vedclass · Accepted Answer

$\frac{82}{3} \,cm$

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(B) Given: Focal length of lens $A$, $f_1 = 2 \,cm$. Focal length of lens $B$, $f_2 = 5 \,cm$. Distance between lenses $d = 14 \,cm$. Object distance for lens $A$, $u_1 = -3 \,cm$.For lens $A$, using the lens formula $\frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f_1}$:$\frac{1}{v_1} - \frac{1}{-3} = \frac{1}{2}$$\frac{1}{v_1} = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}$$v_1 = 6 \,cm$.The image formed by lens $A$ acts as an object for lens $B$. The distance of this image from lens $B$ is $u_2 = -(d - v_1) = -(14 - 6) = -8 \,cm$.For lens $B$, using the lens formula $\frac{1}{v_2} - \frac{1}{u_2} = \frac{1}{f_2}$:$\frac{1}{v_2} - \frac{1}{-8} = \frac{1}{5}$$\frac{1}{v_2} = \frac{1}{5} - \frac{1}{8} = \frac{8 - 5}{40} = \frac{3}{40}$$v_2 = \frac{40}{3} \,cm$.The distance of the final image from lens $A$ is the distance between the lenses plus the distance of the image from lens $B$:Distance $= d + v_2 = 14 + \frac{40}{3} = \frac{42 + 40}{3} = \frac{82}{3} \,cm$.

Two lenses $A$ and $B$ having focal lengths $2.0 \,cm$ and $5.0 \,cm$, respectively, are placed $14 \,cm$ apart. Lens $A$ is placed to the left of lens $B$. An object is placed $3 \,cm$ to the left of lens $A$. The distance of the final image from lens $A$ will be:

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