Question

Since the objective lens merely forms an enlarged real image that is viewed by the eyepiece, the overall angular magnification M of the compound microscope is the product of the lateral magnification $m_1$ of the objective and the angular magnification $M_2$ of the eyepiece. The former is given by
$m_1=\frac{S_1^{{}^{\prime }}}{S_1}$
Where $S_{1}and{S}_1^{{}^{\prime }}$ are the object and image distance for the objective lens. Ordinarily the object is very close to the focus, resulting in an image whose distance from the objective is much larger than the focal length $f_1$. Thus $S_1$ is approximately equal to $f_1$ and $m_1$ =$-\frac{S_1^{{}^{\prime }}}{f_1}$, approximately. The angular magnification of the eyepiece from $M=-\frac{u^{{}^{\prime }}}{u}=\frac{y/f}{y/25}=\frac{25}{f}$ (f in centimeters) is $M_2=25cm/f_2,$ Where $f_2$ is the focal length of the eyepiece, considered as a simple lens. Hence the overall magnification M of the compound microscope is, apart from a negative sign, which is customarily ignored,
$M=m_1{M}_2=\frac{\left(25cm\right)S_1^{{}^{\prime }}}{f}$
The magnifying power given in the paragraph is approximate. What is maximum magnification?

• A. $\frac{V_0}{u_0}(1+\frac{D}{f_e})$
• B. $\frac{V_0}{u_0}(1-\frac{D}{f_e})$
• C. $\frac{V_0}{u_0}(1+\frac{f_e}{D})$
• D. $\frac{V_0}{u_0}(1-\frac{f_e}{D})$

Answer 1

The correct option is A $\frac{V_0}{u_0}(1+\frac{D}{f_e})$

The magnifying power of compound microscope is given by,

$M=m_1{m}_2$ .

By definition, linear magnification by objective is

$m_1=v_o/u_o$

linear magnification by eyepiece is

$m_2=v_e/u_e$

For eyepiece, by lens equation

$\frac{1}{v_e}-\frac{1}{u_e}=\frac{1}{f_e}$

$\frac{v_e}{u_e}=1-\frac{v_e}{f_e}$

Hence, $m_2=1-\frac{v_e}{f_e}$

For maximum magnification, the virtual image should be at $D$ (least distance of distinct vision),

$v_e=-D$

Therefore, $m_2=1+\frac{D}{f_e}$

Maximum magnification,

$M_{max}=\frac{v_o}{u_o}(1+\frac{D}{f_e})$