Search any question & find its solution
Question:
Answered & Verified by Expert
The focal length of objective and eye lens of a microscope are $4 \mathrm{~cm}$ and $8 \mathrm{~cm}$ respectively. If the least distance of distinct vision is $24 \mathrm{~cm}$ and object distance is $4.5 \mathrm{~cm}$ from the objective lens, then the magnifying power of the microscope will be
Options:
Solution:
1983 Upvotes
Verified Answer
The correct answer is:
32
For objective lens $\frac{1}{f_o}=\frac{1}{v_o}-\frac{1}{u_o}$
$\Rightarrow \frac{1}{(+4)}=\frac{1}{v_o}-\frac{1}{(-4.5)} \Rightarrow v_o=36 \mathrm{~cm}$
$\therefore\left|m_D\right|=\frac{v_o}{u_o}\left(1+\frac{D}{f_e}\right)=\frac{36}{4.5}\left(1+\frac{24}{8}\right)=32$
$\Rightarrow \frac{1}{(+4)}=\frac{1}{v_o}-\frac{1}{(-4.5)} \Rightarrow v_o=36 \mathrm{~cm}$
$\therefore\left|m_D\right|=\frac{v_o}{u_o}\left(1+\frac{D}{f_e}\right)=\frac{36}{4.5}\left(1+\frac{24}{8}\right)=32$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.