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

The correct option is B $15cm$
$\begin{array}{c}\text{fobjective}=f_0=1cm\\ \text{feye peice}=f_e=5cm\\ m_2=45m=\text{magnifieng}\\ \text{power for relaxed eye}\\ \Delta =\text{least}D\text{istance of}\\ \text{vision for normal eye}\\ \text{ie}25cm\\ L_{\infty }=\text{Length of tube}\\ \text{for}{m}_{\infty }.\end{array}$

$\begin{array}{c}\text{By using the formula:}\\ m_{\infty }=\frac{(L_{\infty }-f_0-f_c)\times D}{f_0\times f_e}\\ 45=\frac{(L_{\infty }-1-5)\times 25}{1\times 5}\\ 9=L_{\infty }-6\\ L_{\infty }=15cm\end{array}$

Hence option (c) is correct