The focal length of the objective and eye len | vedclass

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Question

(B) For the objective lens, the lens formula is given by $\frac{1}{f_o} = \frac{1}{v_o} - \frac{1}{u_o}$.Given $f_o = 4 \, cm$ and $u_o = -4.5 \, cm$.

Vedclass · Accepted Answer

$32$

Vedclass · Answer

(B) For the objective lens, the lens formula is given by $\frac{1}{f_o} = \frac{1}{v_o} - \frac{1}{u_o}$.Given $f_o = 4 \, cm$ and $u_o = -4.5 \, cm$.Substituting the values: $\frac{1}{4} = \frac{1}{v_o} - \frac{1}{-4.5} \Rightarrow \frac{1}{v_o} = \frac{1}{4} - \frac{1}{4.5} = \frac{4.5 - 4}{18} = \frac{0.5}{18} = \frac{1}{36}$.Thus, $v_o = 36 \, cm$.The magnifying power $M$ for a microscope when the final image is formed at the least distance of distinct vision $D$ is given by $M = \frac{v_o}{|u_o|} \left( 1 + \frac{D}{f_e} ight)$.Given $D = 24 \, cm$ and $f_e = 8 \, cm$.Substituting the values: $M = \frac{36}{4.5} \left( 1 + \frac{24}{8} ight) = 8 	imes (1 + 3) = 8 	imes 4 = 32$.

The focal length of the objective and eye lens of a microscope are $4 \, cm$ and $8 \, cm$ respectively. If the least distance of distinct vision is $24 \, cm$ and the object distance is $4.5 \, cm$ from the objective lens, then the magnifying power of the microscope will be:

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