The focal length of the objective lens of a t | vedclass

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(A) Given: Focal length of objective lens $f_o = 30 	ext{ cm}$, focal length of eye lens $f_e = 3 	ext{ cm}$, and object distance $u_o = -200 	ext{

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$38.3$

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(A) Given: Focal length of objective lens $f_o = 30 	ext{ cm}$, focal length of eye lens $f_e = 3 	ext{ cm}$, and object distance $u_o = -200 	ext{ cm}$ $(2 	ext{ m} = 200 	ext{ cm})$.For the objective lens, using the lens formula $\frac{1}{f_o} = \frac{1}{v_o} - \frac{1}{u_o}$:$\frac{1}{30} = \frac{1}{v_o} - \frac{1}{-200}$$\frac{1}{v_o} = \frac{1}{30} - \frac{1}{200} = \frac{20 - 3}{600} = \frac{17}{600}$$v_o = \frac{600}{17} \approx 35.3 	ext{ cm}$.To see a clear image, the final image should be formed at infinity, which means the intermediate image formed by the objective lens must lie at the focal point of the eye lens.Therefore, the distance between the objective lens and the eye lens is $L = v_o + f_e$.$L = 35.3 	ext{ cm} + 3 	ext{ cm} = 38.3 	ext{ cm}$.

The focal length of the objective lens of a telescope is $30 \text{ cm}$ and that of its eye lens is $3 \text{ cm}$. It is focused on a scale at a distance $2 \text{ m}$ from it. The distance of the objective lens from the eye lens to see the clear image is: (in $\text{ cm}$)

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