
Derive the expression for the angular magnification of a simple microscope.
Answer
482.7k+ views
Hint : When the object is at a distance closer than the distance of the least distinct vision, we can use a microscope to view the object clearly. The angular magnification of a microscope is defined as the ratio of the angle subtended by the image after refraction from the microscope to the angle subtended by the object at an unaided eye/without the apparatus.
Complete step by step answer
The angular magnification is defined as follows:
${\text{M = }}\dfrac{{{\text{angle subtended by eye using instrument}}}}{{{\text{angle subtended at unaided eye}}}}$
When looking at a small object, as shown above, the angle subtended by the object at the unaided eye is very small and can be approximately written as
$\alpha = \dfrac{h}{d}$
When a lens is placed in between the object and the eye as is done in a simple microscope shown above, the angle subtended by the object will be
$\beta = \dfrac{H}{d}$
Since the image will be formed at the least distance of distinct vision $D$, i.e. $d = D$, the angular magnification can be defined as:
$M = \dfrac{\alpha }{\beta } = \dfrac{{H/D}}{{h/D}}$
$ \Rightarrow M = \dfrac{H}{h}$
Now from the lens-maker formula, we can write
$\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$
We can rearrange the above equation to write,
\[\dfrac{v}{f} = 1 + \dfrac{v}{u}\]
Since the ratio of the image to the object position is also equal to magnification and we want the image to form at the least distance of distinct vision $D$ which is typically $25\,cm$, we can say $v/u = M$ and $v = D$ and hence write
$M = \dfrac{D}{f} - 1$
This is the magnification power for a simple microscope to form an image at the least distance of distinct vision of the observer.
Note
If the object is placed at infinity, the angle $\beta $ subtended by the object is the same with or without the lens as. Also, the image formed by the lens will be formed at a distance equal to the focal length of the lens. So $d = f$ and we can write
$\beta = \dfrac{h}{f}$
So,
$M = \dfrac{\alpha }{\beta } = \dfrac{{h/f}}{{h/D}}$
$ \Rightarrow M = \dfrac{D}{f}$
Complete step by step answer
The angular magnification is defined as follows:
${\text{M = }}\dfrac{{{\text{angle subtended by eye using instrument}}}}{{{\text{angle subtended at unaided eye}}}}$

When looking at a small object, as shown above, the angle subtended by the object at the unaided eye is very small and can be approximately written as
$\alpha = \dfrac{h}{d}$

When a lens is placed in between the object and the eye as is done in a simple microscope shown above, the angle subtended by the object will be
$\beta = \dfrac{H}{d}$
Since the image will be formed at the least distance of distinct vision $D$, i.e. $d = D$, the angular magnification can be defined as:
$M = \dfrac{\alpha }{\beta } = \dfrac{{H/D}}{{h/D}}$
$ \Rightarrow M = \dfrac{H}{h}$
Now from the lens-maker formula, we can write
$\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$
We can rearrange the above equation to write,
\[\dfrac{v}{f} = 1 + \dfrac{v}{u}\]
Since the ratio of the image to the object position is also equal to magnification and we want the image to form at the least distance of distinct vision $D$ which is typically $25\,cm$, we can say $v/u = M$ and $v = D$ and hence write
$M = \dfrac{D}{f} - 1$
This is the magnification power for a simple microscope to form an image at the least distance of distinct vision of the observer.
Note
If the object is placed at infinity, the angle $\beta $ subtended by the object is the same with or without the lens as. Also, the image formed by the lens will be formed at a distance equal to the focal length of the lens. So $d = f$ and we can write
$\beta = \dfrac{h}{f}$
So,
$M = \dfrac{\alpha }{\beta } = \dfrac{{h/f}}{{h/D}}$
$ \Rightarrow M = \dfrac{D}{f}$
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Basicity of sulphurous acid and sulphuric acid are

Master Class 9 General Knowledge: Engaging Questions & Answers for Success

Trending doubts
Which one of the following is a true fish A Jellyfish class 12 biology CBSE

Draw a labelled sketch of the human eye class 12 physics CBSE

a Tabulate the differences in the characteristics of class 12 chemistry CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Why is the cell called the structural and functional class 12 biology CBSE

What are the major means of transport Explain each class 12 social science CBSE
