In a compound microscope,the focal lengths of | vedclass

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(D) For the objective lens:Given: $u_o = -2 	ext{ cm}$,$f_o = 1.5 	ext{ cm}$.Using the lens formula $\frac{1}{v_o} - \frac{1}{u_o} = \frac{1}{f_o}$:

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$11$

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(D) For the objective lens:Given: $u_o = -2 	ext{ cm}$,$f_o = 1.5 	ext{ cm}$.Using the lens formula $\frac{1}{v_o} - \frac{1}{u_o} = \frac{1}{f_o}$:$\frac{1}{v_o} - \frac{1}{-2} = \frac{1}{1.5}$$\frac{1}{v_o} = \frac{1}{1.5} - \frac{1}{2} = \frac{2-1.5}{3} = \frac{0.5}{3} = \frac{1}{6}$So,$v_o = 6 	ext{ cm}$.For the eyepiece lens:Given: $v_e = -25 	ext{ cm}$ (final image is virtual),$f_e = 6.25 	ext{ cm}$.Using the lens formula $\frac{1}{v_e} - \frac{1}{u_e} = \frac{1}{f_e}$:$\frac{1}{-25} - \frac{1}{u_e} = \frac{1}{6.25}$$\frac{1}{u_e} = -\frac{1}{25} - \frac{1}{6.25} = -\frac{1}{25} - \frac{4}{25} = -\frac{5}{25} = -\frac{1}{5}$So,$u_e = -5 	ext{ cm}$.The distance between the two lenses is $L = |v_o| + |u_e| = 6 	ext{ cm} + 5 	ext{ cm} = 11 	ext{ cm}$.

In a compound microscope,the focal lengths of two lenses are $1.5 \text{ cm}$ and $6.25 \text{ cm}$. An object is placed at $2 \text{ cm}$ from the objective and the final image is formed at $25 \text{ cm}$ from the eye lens. The distance between the two lenses is .............. (in $\text{cm}$).

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