A small object is placed at a distance of 4 , | vedclass

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(B) For the objective lens, the object distance $u_o = -4 \,m$ and focal length $f_o = 2 \,m$. Using the lens formula $\frac{1}{v_o} - \frac{1}{u_o} =

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will be real

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(B) For the objective lens, the object distance $u_o = -4 \,m$ and focal length $f_o = 2 \,m$. Using the lens formula $\frac{1}{v_o} - \frac{1}{u_o} = \frac{1}{f_o}$:$\frac{1}{v_o} - \frac{1}{-4} = \frac{1}{2} \Rightarrow \frac{1}{v_o} = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$.So, $v_o = 4 \,m$. This image acts as an object for the eyepiece.Let the distance between the objective and the eyepiece be $L$. The object distance for the eyepiece is $u_e = -(4 - L)$.Using the lens formula for the eyepiece: $\frac{1}{v_e} - \frac{1}{u_e} = \frac{1}{f_e}$.$\frac{1}{v_e} - \frac{1}{-(4 - L)} = \frac{1}{0.2} \Rightarrow \frac{1}{v_e} = 5 - \frac{1}{4 - L}$.Since $L$ is typically around $f_o + f_e = 2.2 \,m$, $4 - L$ is positive and less than $4$. Thus, $\frac{1}{4 - L} > 0.25$. Therefore, $\frac{1}{v_e} = 5 - (	ext{a value greater than } 0.25)$, which is positive. Since $v_e$ is positive, the final image is real.

$A$ small object is placed at a distance of $4 \,m$ from the objective of a telescope of focal length $2 \,m$. The focal length of the eyepiece is $0.2 \,m$. The final image of the object

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