The focal lengths of the objective and the ey | vedclass

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(A) Given: Focal length of objective $f_o = 2 \,cm$, focal length of eyepiece $f_e = 3 \,cm$, and distance between lenses $L = 15 \,cm$.Since the fina

Vedclass · Accepted Answer

$2.4 \,cm, 12 \,cm$

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(A) Given: Focal length of objective $f_o = 2 \,cm$, focal length of eyepiece $f_e = 3 \,cm$, and distance between lenses $L = 15 \,cm$.Since the final image is formed at infinity, the image formed by the objective lens must lie at the principal focus of the eyepiece.Therefore, the image distance from the eyepiece is $v_e = f_e = 3 \,cm$.The image distance from the objective lens is $v_o = L - f_e = 15 \,cm - 3 \,cm = 12 \,cm$.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} = \frac{1}{12} - \frac{1}{u_o}$$\frac{1}{u_o} = \frac{1}{12} - \frac{1}{2} = \frac{1 - 6}{12} = -\frac{5}{12}$$u_o = -\frac{12}{5} = -2.4 \,cm$.The magnitude of the object distance is $2.4 \,cm$ and the image distance is $12 \,cm$.

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