Magnification of a compound microscope is 30. | vedclass

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Question

(B) The total magnification $m$ of a compound microscope when the final image is formed at the near point $D$ is given by the formula:$m = m_{o} 	ime

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(B) The total magnification $m$ of a compound microscope when the final image is formed at the near point $D$ is given by the formula:$m = m_{o} 	imes m_{e}$where $m_{o}$ is the magnification of the objective lens and $m_{e}$ is the magnification of the eyepiece.For an eyepiece,the magnification when the image is formed at the near point is given by $m_{e} = (1 + \frac{D}{f_{e}})$.Given:Total magnification $m = 30$Focal length of eyepiece $f_{e} = 5 \, cm$Distance of distinct vision $D = 25 \, cm$Substituting the values into the formula:$30 = m_{o} 	imes (1 + \frac{25}{5})$$30 = m_{o} 	imes (1 + 5)$$30 = m_{o} 	imes 6$$m_{o} = \frac{30}{6} = 5$Therefore,the magnification of the objective lens is $5$.

Magnification of a compound microscope is $30$. The focal length of the eyepiece is $5 \, cm$ and the image is formed at the near point (distance of distinct vision) of $25 \, cm$. The magnification of the objective lens is:

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