Question

In a compound microscope, the final image is formed at a distance ${}^{\prime }{d}^{\prime }$ from the eyepiece, whose focal length is $f_e$, where $d>D$. The object is placed at a distance ${}^{\prime }{u}^{\prime }$ from the objective lens, whose focal length is $f_o$. The angular magnification of the microscope is

• A. $\frac{f_o}{u-f_o}\times \left(\frac{D}{d}+\frac{1}{f_e}\right)$
• B. $\frac{D{f}_o}{u-f_o}\times \left(\frac{1}{d}+\frac{1}{f_e}\right)$
• C. $\frac{D{f}_o}{u-f_o}\times \left(\frac{1}{d}+\frac{1}{f_o}\right)$
• D. $\frac{D{f}_o}{u+f_o}\times \left(\frac{1}{d}+\frac{1}{f_e}\right)$

Answer 1

The correct option is D $\frac{D{f}_o}{u+f_o}\times \left(\frac{1}{d}+\frac{1}{f_e}\right)$

$angula{r}_{\text{magnification}}=\frac{1}{latera{l}_{\text{magnification}}}$

$angula{r}_{\text{magnification}}=\frac{D{f}_o}{u+f_o}\times (\frac{1}{d}+\frac{1}{f_e})$

option $D$ is correct