Question

The focal length of the objective and eye piece of a compound microscope are $1 \mathrm{cm}$ and $5 \mathrm{cm}$ respectively. An object is placed at a distance of $1.1 \mathrm{cm}$ from objective. If final image is formed by at least distance vision, the magnifying power is:
(A) 60
(B) 50
(C) 40
(D) None of these

Answer 1

Hint To solve this question, it is required to know that compound microscope is an upright microscope that uses two sets of lenses (a compound lens system) to obtain higher magnification than a stereo microscope. A compound microscope provides a two-dimensional image, while a stereo microscope provides a three-dimensional image. The difference is a simple microscope has one lens where a compound microscope has an objective lens and an eyepiece with a longer focal length. A simple microscope only has one type of lens, usually objective lens, but a compound microscope has both an objective lens and ocular lens. The term compound in compound microscopes refers to the microscope having more than one lens. Devised with a system of combination of lenses, a compound microscope consists of two optical parts, namely the objective lens and the ocular lens.

Complete step by step answer
Given that,
Focal length of objective $\mathrm{f}_{\mathrm{o}}=1 \mathrm{cm}$
Focal length of eyepiece $\mathrm{f}_{\mathrm{e}}=5 \mathrm{cm}$
Distance of the object ${{u}_{o}}=1.1\text{cm}$
Distinct vision $=\mathrm{D}$
Now, the magnifying power of a compound microscope is the product of the linear magnification of the objective and the magnifying power of the eyepiece.
$\mathrm{M} . \mathrm{P}=\mathrm{M}_{\mathrm{o}} \times \mathrm{M}_{\mathrm{e}}$
$\text{M}.\text{P}=\left( \dfrac{{{\text{v}}_{o}}}{{{\text{v}}_{\text{e}}}} \right)\times \left( \dfrac{\text{D}}{{{\text{u}}_{o}}} \right)\ldots ..(\text{I})$
Now, for the objective lens $\dfrac{1}{\text{f}}=\dfrac{1}{{{\text{v}}_{o}}}-\dfrac{1}{{{\text{u}}_{o}}}$
${{\text{v}}_{\text{o}}}=\dfrac{{{\text{u}}_{o}}{{\text{f}}_{\text{o}}}}{{{\text{u}}_{\text{o}}}+{{\text{f}}_{\text{o}}}}$
$\dfrac{{{v}_{o}}}{{{u}_{o}}}=\dfrac{{{f}_{o}}}{{{u}_{o}}+{{f}_{o}}}$
Now, put the value in equation (I) $\text{M}.\text{P}=\left( \dfrac{{{\text{f}}_{\text{o}}}}{{{\text{u}}_{\text{o}}}+{{\text{f}}_{\text{o}}}} \right)\times \left( \dfrac{\text{D}}{{{\text{u}}_{o}}} \right)$
Now, because ${{u}_{o}}>{{f}_{o}}$
So, $\mathrm{M.P}=-\left(\dfrac{\mathrm{f}_{\mathrm{o}}}{\mathrm{u}_{\mathrm{o}}+\mathrm{f}_{\mathrm{o}}}\right) \times\left(\dfrac{\mathrm{D}}{\mathrm{u}_{\mathrm{o}}}\right)$
Now, the distance of distinct vision, $\mathrm{D}$ may be taken as $25 \mathrm{cm}$.
So, $\mathrm{M.P}=-\left(\dfrac{\mathrm{f}_{\mathrm{o}}}{\mathrm{u}_{\mathrm{o}}+\mathrm{f}_{\mathrm{o}}}\right) \times\left(\dfrac{\mathrm{D}}{\mathrm{u}_{\mathrm{o}}}\right)$
M.P $=-\left(\dfrac{1}{1.1+1}\right) \times \dfrac{25}{5}$
$\mathrm{M} \cdot \mathrm{P}=-50$
The negative sign indicates that the image is inverted Hence, the magnifying power is 50.

Therefore, the correct answer is Option B.

Note We need to know that the term focal length refers to the amount of distance required between the objective lens and the top of our object, in order to be able to view an image through the microscope that is in-focus. When using a biological microscope, the higher our objective magnification, the shorter the focal length generally is. The focal length of a microscope objective is typically between 2 mm and 40 mm. However, that parameter is often considered as less important, since magnification and numerical aperture are sufficient for quantifying the essential performance in a microscope. The objective lens of a microscope is the one at the bottom near the sample. At its simplest, it is a very high-powered magnifying glass, with very short focal length. This is brought very close to the specimen being examined so that the light from the specimen comes to a focus inside the microscope tube.