A microscope has an objective of focal length | vedclass

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(B) Given:Objective focal length $f_o = 2 \ cm$Eyepiece focal length $f_e = 4 \ cm$Tube length $L = 40 \ cm$Distance of distinct vision $D = 25 \ cm$F

Vedclass · Accepted Answer

$125$

Vedclass · Answer

(B) Given:Objective focal length $f_o = 2 \ cm$Eyepiece focal length $f_e = 4 \ cm$Tube length $L = 40 \ cm$Distance of distinct vision $D = 25 \ cm$For a compound microscope,the magnifying power $m$ when the final image is formed at the near point (distance of distinct vision) is given by the formula:$m = \frac{L}{f_o} 	imes \left( 1 + \frac{D}{f_e} ight)$However,in many textbook contexts for this specific problem type where $L$ is defined as the distance between the focal points,the formula simplifies to:$m = \frac{L}{f_o} 	imes \frac{D}{f_e}$Substituting the values:$m = \frac{40}{2} 	imes \frac{25}{4}$$m = 20 	imes 6.25$$m = 125$

$A$ microscope has an objective of focal length $2 \ cm$,an eyepiece of focal length $4 \ cm$,and a tube length of $40 \ cm$. If the distance of distinct vision is $25 \ cm$,the magnification of the microscope is:

Similar Questions

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