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:

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

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

 Hence option (c) is correct