Question

Find the maximum magnifying power of a compound microscope having a 25 diopter lens as the objective, a 5 diopter lens as the eyepiece, and the separation 30 cm between the two lenses. The least distance for a clear vision is 25 cm.
A) $8.4$
B) $7.4$
C) $9.4$
D) $10.4$

Answer 1

Hint
The focal length of a lens is the inverse of the dioptre of the lens. We need to calculate the focal length of the object lens and the eyepiece and then the position of the final image formed in a compound microscope and then we can use the maximum magnifying power formula to find the answer..
Formula used: In this solution we will be using the following formula,
Lens formula: $\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$ where $v$ is the image distance, $u$ is the object distance, and $f$ is the focal length of the lens
Maximum magnifying power: $m = \dfrac{{ - v}}{u}\left[ {1 + \dfrac{D}{{{f_e}}}} \right]$ where $D$ is the least distance for clear vision, ${f_e}$ is the focal length of the eyepiece.

Complete step by step answer
For a lens, the focal length can be calculated as the inverse of the magnifying power of the lens. Thus for the eyepiece, the focal length will be
$\Rightarrow {f_e} = \dfrac{1}{5} = 0.2$
And the focal length for the objective lens will be
$\Rightarrow {f_o} = \dfrac{1}{{25}} = 04$
Now using the lens formula, let us find the position of the final image after refractions from the two lenses. After refraction with the eyepiece, the image must be formed at the eye of the observer which is at a distance of ${v_e} = 25\,cm$ from the eyepiece. So, using the lens formula for the eyepiece, we can write
$\Rightarrow \dfrac{1}{{{v_e}}} - \dfrac{1}{{{u_e}}} = \dfrac{1}{{{f_e}}}$
Substituting the values we get
$\Rightarrow \dfrac{1}{{ - 25}} - \dfrac{1}{{{u_e}}} = \dfrac{1}{{0.2}}$
Solving for ${u_e}$, we can write
$\Rightarrow {u_e} = 11.11\,cm$
Now the object for the eyepiece is actually the image formed by the objective lens. Since the separation between the two lenses is 30 cm, we can find the distance of the image of the objective lens from it as
$\Rightarrow {v_0} = 30 - 11.11$
Which is equal to,
$\Rightarrow {v_0} = 18.89\,cm$
Again, using the lens maker formula for the objective lens, we can write
$\Rightarrow \dfrac{1}{{{v_o}}} - \dfrac{1}{{{u_0}}} = \dfrac{1}{{{f_0}}}$
Substituting the values we get,
$\Rightarrow \dfrac{1}{{ - 18.89}} - \dfrac{1}{{{u_0}}} = \dfrac{1}{4}$
Solving for ${u_0}$, we get
$\Rightarrow {u_o} = - 5.07\,cm$
We can now find the maximum magnifying power of the lens using the formulas
$\Rightarrow m = \dfrac{{ - {v_o}}}{{{u_o}}}\left[ {1 + \dfrac{D}{{{f_e}}}} \right]$
$\Rightarrow m = \dfrac{{ - 18.89}}{{ - 5.07}}\left[ {1 + \dfrac{{25}}{{20}}} \right]$
Solving for $m$, we get
$m = 8.376 \approx 8.4$ which corresponds to (A).

Note
We must be careful in substituting the proper sign corresponding to the object, image position, and the focal length of the lens depending on the kind of lens being used. The magnifying power of the system depends on the image and object position for the objective lens and on the focal length of the eyepiece which must be carefully substituted to find the correct option.