Magnifying Power of Simple Microscope

Simple microscope enables us to see an enlarged image of an object

Often we have seen jewellers using a simple microscope to see a magnified view of fine parts of a jewellery

It uses a lens to enlarge an object (up-to $200X$) through angular magnification alone

Let us see how a simple microscope forms an enlarged image.

The apparent size of an object depends upon its visual angle subtended on the eye

The greater the angle subtended by the final image on the eye, the greater is the apparent size of the object

In order to increase the visual angle, the object distance must be small. But the eye cannot see an object clearly if it is closer than 25 cm

This distance is called the least distance of distinct vision. It is denoted by D

To solve this problem, simple microscope uses a convex lens of short focal length, to see the details of an object within the least distance of distinct vision

Magnifying power of a simple microscope

The above expression gives the magnifying power of a simple microscope

Let the angle subtended by the object when placed at a distance D (Least Distance of Distinct Vision) from the eye be $\alpha$

Let us consider an object $OP$ placed at a distance $u$ within the focal length of a microscope such that a magnified image $IQ$ is produced at $D$

Let the angle subtended by the image when seen through the microscope be $\beta$

$\text{The magnifying power},M=\frac{\beta }{\alpha }$ $$\begin{gathered} \text{Since both }\alpha \&\beta \text{ are small;} \\ \text{by tan approximation rule} \\ \text{ we can write-} \\ M=\frac{\tan \beta }{\tan \alpha } \end{gathered}$$

$\tan \alpha =\frac{AB}{BE}=\frac{AB}{D}$

And, $\tan \beta =\frac{OP}{PE}=\frac{OP}{u}$

Now we know that, OP=AB= Height of object $\therefore \tan \beta =\frac{OP}{u}=\frac{AB}{u}$

Let us now use the values of $\alpha \&\beta$ in the formula of M

$M=\frac{\tan \beta }{\tan \alpha }=\frac{\frac{AB}{u}}{\frac{AB}{D}}=\frac{D}{u}$

Expressing the magnifying power in terms of the focal length of the lens used

$$\begin{gathered} \text{We know the lens formula-} \\ \frac{1}{f}=\frac{1}{u}+\frac{1}{v} \\ \text{where,} \\ \text{f=Focal Length of the Lens} \\ \text{u=Object distance} \\ \text{v=Image Distance} \end{gathered}$$

$$\begin{gathered} \text{Using the necessary sign} \\ \text{ conventions,} \\ f=+f \\ u=+u \\ v=-D \end{gathered}$$

$$\begin{gathered} \text{Applying the lens formula-} \\ \frac{1}{f}=\frac{1}{u}+\frac{1}{v} \\ \to \frac{1}{f}=\frac{1}{u}+\frac{1}{-D} \\ \to \frac{1}{f}=\frac{1}{u}-\frac{1}{D} \end{gathered}$$

$$\begin{gathered} \text{Multiplying both sides by D} \\ \to \frac{D}{f}=\frac{D}{u}-\frac{D}{D} \\ \to \frac{D}{f}=\frac{D}{u}-1 \\ \to \frac{D}{f}+1=\frac{D}{u} \end{gathered}$$

$$\begin{gathered} \frac{D}{f}+1=\frac{D}{u}; \\ \text{But we know the term: }\frac{D}{u}=M \\ \therefore \text{We can write-}\frac{D}{f}+1=M \end{gathered}$$

$$\begin{gathered} \text{The magnifying power of a} \\ \text{simple microscope, M=}\frac{D}{f}+1 \end{gathered}$$

Revision

Simple microscopes are devices which uses lens to form an enlarged image of an object

$$\begin{gathered} \text{The magnifying power of a} \\ \text{simple microscope, M=}\frac{D}{f}+1 \\ \text{where,} \\ \text{D= Least Distance of Distinct } \\ \text{Vision} \\ \text{f= Focal length of the lens used} \end{gathered}$$

The End