The focal lengths of the objective and the ey | vedclass

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(A) Given: Focal length of objective $f_o = 2.0 \, cm$,focal length of eye-piece $f_e = 3.0 \, cm$,and tube length $L = 15.0 \, cm$.When the final ima

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$2.4$ and $12.0$

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(A) Given: Focal length of objective $f_o = 2.0 \, cm$,focal length of eye-piece $f_e = 3.0 \, cm$,and tube length $L = 15.0 \, cm$.When the final image is formed at infinity,the eye-piece is adjusted such that the intermediate image formed by the objective lies at the principal focus of the eye-piece.Therefore,the distance of the intermediate image from the objective $(v_o)$ is given by $L = v_o + f_e$.$15.0 = v_o + 3.0 \Rightarrow v_o = 12.0 \, cm$.Now,using the lens formula for the objective lens: $\frac{1}{f_o} = \frac{1}{v_o} - \frac{1}{u_o}$.Substituting the values: $\frac{1}{2.0} = \frac{1}{12.0} - \frac{1}{u_o}$.$\frac{1}{u_o} = \frac{1}{12.0} - \frac{1}{2.0} = \frac{1 - 6}{12.0} = -\frac{5}{12.0}$.$u_o = -\frac{12.0}{5} = -2.4 \, cm$.The magnitude of the object distance is $2.4 \, cm$ and the image distance is $12.0 \, cm$.

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