Question

If we need a magnification of 375 from a compound microscope of tube length 150mm and an objective of focal length 5mm, the focal length of the eyepiece should be close to
(A) $2mm$
(B) $12mm$
(C) $33mm$
(D) $22mm$

Answer 1

Hint: The focal length of the eyepiece can be determined by using the magnification of the compound microscope formula. By using the given information in the question, the focal length of the eyepiece in terms of the millimeter can be determined.

Formula used:
Magnification of the compound microscope,
$M={\displaystyle \frac{v_0}{u_0}}\left(1+{\displaystyle \frac{d}{f_e}}\right)$
Where, $M$ is the magnification, $v_0$ is the length of the tube, $u_0$ is the object distance from the lens, $d$ is the diameter of the microscope lens and $f_e$ is the focal length of the eye piece.

Complete step by step answer:
Given that,
The magnification of the microscope, $M=375$.
The length of the microscope tube, $v_0=150mm$.
The object distance from the lens, $u_0=5mm$
The diameter of the compound microscope, $d=250mm$.
Magnification of the compound micro scope,
$\Rightarrow M={\displaystyle \frac{v_0}{u_0}}\left(1+{\displaystyle \frac{d}{f_e}}\right)......................\left(1\right)$
By substituting the magnification of the microscope, the length of the microscope tube, the object distance from the lens, and the diameter of the compound microscope, in the equation (1), then
$\Rightarrow 375={\displaystyle \frac{150}{5}}\left(1+{\displaystyle \frac{250}{f_e}}\right)$
By taking the term ${\displaystyle \frac{150}{5}}$ from the RHS to LHS, then the above equation is written as,
$\Rightarrow {\displaystyle \frac{375\times 5}{150}}=\left(1+{\displaystyle \frac{250}{f_e}}\right)$
By solving the terms in the LHS, then the above equation is written as,
$\Rightarrow 12.5=1+{\displaystyle \frac{250}{f_e}}$
By taking the term $1$ from RHS to LHS, then
$\Rightarrow 12.5-1={\displaystyle \frac{250}{f_e}}$
By solving the terms in the LHS, then the above equation is written as,
$\Rightarrow 11.5={\displaystyle \frac{250}{f_e}}$
By keeping the term $f_e$ on one side and the other terms on another side, then
$\Rightarrow f_e={\displaystyle \frac{250}{11.5}}$
On dividing the above equation, then the above equation is written as,
$\Rightarrow f_e=21.7mm$
Then, the above equation is written as,
$\Rightarrow f_e\simeq 22mm$
Thus, the above equation shows the focal length of the eye piece.

Hence option (D) is the correct answer.

Note:
The diameter of the microscope lens is not given in the question. So, let us take the standard lens diameter of the microscope. Thus, the final answer shows, the distance of the lens which is placed in front of the eye, that distance is called the focal length of the eyepiece.