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(B) The magnifying power of a compound microscope when the final image is formed at infinity is given by the formula: $m = -\frac{v_o}{u_o} 	imes \fr

Vedclass · Accepted Answer

$200$

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(B) The magnifying power of a compound microscope when the final image is formed at infinity is given by the formula: $m = -\frac{v_o}{u_o} 	imes \frac{D}{f_e}$.Given:Focal length of objective,$f_o = 1.2 \, cm$.Focal length of eye lens,$f_e = 3 \, cm$.Object distance,$u_o = -1.25 \, cm$.Near point distance,$D = 25 \, cm$.Using the lens formula for the objective lens: $\frac{1}{f_o} = \frac{1}{v_o} - \frac{1}{u_o}$.$\frac{1}{1.2} = \frac{1}{v_o} - \frac{1}{-1.25}$.$\frac{1}{v_o} = \frac{1}{1.2} - \frac{1}{1.25} = \frac{1.25 - 1.2}{1.5} = \frac{0.05}{1.5} = \frac{1}{30}$.So,$v_o = 30 \, cm$.Now,calculate the magnifying power:$m = -\left(\frac{30}{-1.25}ight) 	imes \left(\frac{25}{3}ight)$.$m = 24 	imes 8.33 = 200$.Thus,the magnifying power is $200$.

If the focal length of the objective and eye lens are $1.2 \, cm$ and $3 \, cm$ respectively,and the object is placed $1.25 \, cm$ away from the objective lens,and the final image is formed at infinity,the magnifying power of the microscope is:

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