The focal lengths of the objective and eye-le | vedclass

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(C) For a compound microscope,the magnifying power $(M)$ for a relaxed eye (final image at infinity) is given by the formula: $M = \frac{L}{f_o} 	ime

Vedclass · Accepted Answer

$15$

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(C) For a compound microscope,the magnifying power $(M)$ for a relaxed eye (final image at infinity) is given by the formula: $M = \frac{L}{f_o} 	imes \frac{D}{f_e}$ where $L$ is the tube length,$f_o = 1 \,cm$ is the focal length of the objective,$f_e = 5 \,cm$ is the focal length of the eyepiece,and $D = 25 \,cm$ is the least distance of distinct vision. Given $M = 45$,we substitute the values: $45 = \frac{L}{1} 	imes \frac{25}{5}$ $45 = L 	imes 5$ $L = \frac{45}{5} = 9 \,cm$. Wait,checking the standard formula for tube length $L = v_o + f_e$. Since $M = m_o 	imes m_e = (v_o/u_o) 	imes (D/f_e)$,and $1/v_o - 1/u_o = 1/f_o$,for relaxed eye $u_o \approx f_o$,so $v_o \approx L - f_e$. Using $M = \frac{L - f_e}{f_o} 	imes \frac{D}{f_e}$: $45 = \frac{L - 5}{1} 	imes \frac{25}{5}$ $45 = (L - 5) 	imes 5$ $9 = L - 5$ $L = 14 \,cm$. Re-evaluating the provided solution logic: The formula used in the prompt is $M = \frac{(L - f_o - f_e)D}{f_o f_e}$. $45 = \frac{(L - 1 - 5) 	imes 25}{1 	imes 5} \Rightarrow 45 = (L - 6) 	imes 5 \Rightarrow 9 = L - 6 \Rightarrow L = 15 \,cm$.

The focal lengths of the objective and eye-lens of a microscope are $1 \,cm$ and $5 \,cm$ respectively. If the magnifying power for the relaxed eye is $45$,then the length of the tube is $....... \,cm$.

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