In a compound microscope,the focal length of | vedclass

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(A) Given: Focal length of objective $f_o = 1.2 \ cm$,focal length of eyepiece $f_e = 3.0 \ cm$,and object distance $u_o = -1.25 \ cm$. For the object

Vedclass · Accepted Answer

$200$

Vedclass · Answer

(A) Given: Focal length of objective $f_o = 1.2 \ cm$,focal length of eyepiece $f_e = 3.0 \ cm$,and object distance $u_o = -1.25 \ cm$. For the objective lens,using the lens formula $\frac{1}{v_o} - \frac{1}{u_o} = \frac{1}{f_o}$: $\frac{1}{v_o} - \frac{1}{-1.25} = \frac{1}{1.2} \implies \frac{1}{v_o} = \frac{1}{1.2} - \frac{1}{1.25} = \frac{1.25 - 1.2}{1.5} = \frac{0.05}{1.5} = \frac{1}{30}$. So,$v_o = 30 \ cm$. The magnification of the objective is $m_o = \frac{v_o}{u_o} = \frac{30}{-1.25} = -24$. Since the final image is at infinity,the magnification of the eyepiece is $m_e = \frac{D}{f_e} = \frac{25}{3.0} = 8.33$. However,in standard textbook problems for compound microscopes where the final image is at infinity,the formula used is $M = m_o 	imes m_e = (\frac{v_o}{u_o}) 	imes (\frac{D}{f_e})$. Using $M = \frac{v_o}{u_o} 	imes \frac{25}{f_e} = -24 	imes \frac{25}{3} = -24 	imes 8.33 = -200$. The magnitude of magnifying power is $200$.

In a compound microscope,the focal length of the objective lens is $1.2 \ cm$ and the focal length of the eyepiece is $3.0 \ cm$. When the object is kept at $1.25 \ cm$ in front of the objective,the final image is formed at infinity. The magnifying power of the compound microscope is:

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