The focal length of the objective and eye len | vedclass

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(B) For a microscope in normal adjustment,the final image is formed at infinity. The magnification $m$ is given by $m = m_o 	imes m_e$.For the object

Vedclass · Accepted Answer

$110$

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(B) For a microscope in normal adjustment,the final image is formed at infinity. The magnification $m$ is given by $m = m_o 	imes m_e$.For the objective lens,the image is formed at the focal point of the eyepiece. The tube length $L$ is the distance between the focal point of the objective and the focal point of the eyepiece.Given: $f_o = 1.6 \, cm$,$f_e = 2.5 \, cm$,and the distance between lenses $d = 21.7 \, cm$.The distance between the lenses is $d = v_o + f_e$,where $v_o$ is the image distance of the objective.$v_o = d - f_e = 21.7 - 2.5 = 19.2 \, cm$.Magnification of the objective $m_o = \frac{v_o}{u_o}$. Using the lens formula $\frac{1}{f_o} = \frac{1}{v_o} - \frac{1}{u_o}$,we get $\frac{1}{u_o} = \frac{1}{v_o} - \frac{1}{f_o} = \frac{1}{19.2} - \frac{1}{1.6} = \frac{1 - 12}{19.2} = -\frac{11}{19.2}$.So,$m_o = v_o 	imes (\frac{1}{v_o} - \frac{1}{f_o}) = 19.2 	imes (-\frac{11}{19.2}) = -11$.The magnification of the eyepiece $m_e = \frac{D}{f_e}$. Assuming the near point $D = 25 \, cm$,$m_e = \frac{25}{2.5} = 10$.Total magnification $m = |m_o| 	imes m_e = 11 	imes 10 = 110$.

The focal length of the objective and eye lenses of a microscope are $1.6 \, cm$ and $2.5 \, cm$ respectively. The distance between the two lenses is $21.7 \, cm$. If the final image is formed at infinity,what is the linear magnification?

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