Microscopes

It is a tool that helps us see tiny details of objects that we can't view with the naked eye. It's crucial in fields like biology, medicine, and materials science, where it's used to examine the structure and behavior of cells, tissues, and microorganisms.

1.0Visual Angle

• It is the angle, a viewed object or image subtends at the eye. It is also called the object’s angular size.
• Nearer the object larger the visual angle then the object appears big in size.

Note:Near point (N.P.) of normal eye D = 25cm

Far point (F.P.) of normal eye = $\infty$

2.0Angular Magnification (Magnifying Power)

• The magnifying power of a simple microscope is stated as the ratio of the angular size of the image to that of the object when both are viewed at the eye's least distance of distinct vision.

Note:Least distance of distinct vision (typically 25 cm).

$M.P=\frac{visualangleofimageinthepresenceofinstrument\left(\beta \right)}{maximumvisualangleofanobjectintheabsenceofinstrument\left(\alpha \right)}$

3.0Simple Microscope

• A converging lens (convex lens) with a short focal length.
• When the object is positioned between the focal point and the optical center, a virtual, magnified, and upright image is produced.
• It is also referred to as a magnifier.

$Tan\beta =\frac{h_0}{u}=\frac{h_i}{v}$

If angle is small then,     

$\beta =\frac{h_0}{u}=\frac{h_i}{v}$

4.0Calculation Of Magnifying Power

$Tan\alpha \approx \alpha =\frac{h_0}{D}$

$M.P=\frac{\beta }{\alpha }=\frac{\frac{h_0}{u}}{\frac{h_0}{D}}=\frac{D}{|u|}$

$M.P=\frac{D}{|u|}$

Case (1): When image formed at a distance of least distance of distinct vision from the lens.

$\frac{1}{-D}-\frac{1}{-u}=\frac{1}{f}\Rightarrow \frac{1}{u}=\frac{1}{D}+\frac{1}{f}$

$[v=-D\text{ and }u=-ve]$

Multiplying both the sides by D

$\frac{D}{u}=1+\frac{D}{f}\Rightarrow M.P=\frac{D}{u}=1+\frac{D}{f}$

$(M.P{)}_{max}=\frac{D}{u_{min}}=1+\frac{D}{f}$

Case (2): When image formed at infinite distance from the lens.

From lens equation

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\Rightarrow \frac{1}{-\infty }-\frac{1}{-u}=\frac{1}{f}\Rightarrow u=f$

$\text{So, }M.P=\frac{D}{u}=\frac{D}{f}$

$(M.P{)}_{min}=\frac{D}{u_{max}}=\frac{D}{f}$

Example-1.Focal length of a converging lens is 2.5 cm. Find its maximum and minimum magnifying power when used as a simple microscope.

Solution:

$(M.P{)}_{max}=1+\frac{D}{f}=1+\frac{25}{2.5}=11$

$(M.P{)}_{min}=\frac{D}{f}=\frac{25}{2.5}=10$

Example-2.A person with normal vision is reading a book with fine  print using a magnifying glass with a focal length of 5 cm. What are the closest and farthest distances at which the lens should be held from the print for comfortable reading?

Solution:

$\frac{D}{u_{min}}=1+\frac{D}{f}$

$\frac{25}{u_{min}}=1+\frac{25}{5}$

$u_{min}=\frac{25}{6}\text{ cm}$

$\frac{D}{u_{max}}=\frac{D}{f}$

$u_{max}=f$

$u_{max}=5\text{ cm}$

Example-3.A 10 D lens is used as a magnifier. Where should the object be placed to obtain maximum angular magnification for a normal eye (Near Point = 25cm) ?

Solution:

$P=\frac{1}{f}=10$

$f=0.1\text{ m}=10\text{ cm}$

$(M.P{)}_{max}=1+\frac{D}{f}=1+\frac{25}{10}=3.5$

$M.P=\frac{D}{|u|}\Rightarrow 3.5=\frac{25}{u}\Rightarrow u=\frac{50}{7}\text{ cm}$

Alternate Method:

We get maximum M.P where using image formed at least distance of district vision

$v=-25\text{ cm}$

$-\frac{1}{25}-\frac{1}{u}=\frac{1}{10}\Rightarrow u=-\frac{50}{7}\text{ cm}$

5.0Compound microscope

• A compound microscope is a high-magnification laboratory instrument with multiple lenses, used to study the detailed structure of cells, tissues, or organ sections. It can magnify small objects up to 1000 times.

Ray Diagram

• A compound microscope is used to get a more magnified image as compared to a simple microscope. The object is placed in front of the objective lens and the image is seen through the eyepiece.
• The aperture of an objective lens is less as compared to an eye piece because the object is very near so collection of more light is not required.
•  Generally, an object is placed between F - 2F due to this a real, inverted and magnified image is formed between $2F-\infty$. It is known as intermediate image (I). The intermediate image acts as an object for the eyepiece.
• Now the distance between both lenses are adjusted in such a way that the intermediate image falls between the optical centre of the eye piece and its focus. In this condition, the final image $(I_f)$ is virtual, inverted and magnified.

6.0Calculation Of Magnifying Power

$\text{Total magnifying power}==\text{Linear magnification of objective lens}\times \text{angular magnification }MP\text{ of eye}$

$M.P=m_0\times m_e$

$M.P=-\left|\frac{v_0}{u_0}\frac{D}{u_e}\right|$

Case (1): When final image formed at least distance of distinct vision.

$M.P=\frac{v_0}{u_0}\left[1+\frac{D}{f_e}\right]=\frac{f_0}{(f_0+u_0)}\left[1+\frac{D}{f_e}\right]=\frac{f_0-v_0}{f_0}\left[1+\frac{D}{f_e}\right]$

Distance between both the lenses, $L=v_0+|u_e|$

Case (2): When the final image formed at infinity.

$M.P=\frac{v_0}{u_0}\left[\frac{D}{f_e}\right]=\frac{f_0}{(f_0+u_0)}\left[\frac{D}{f_e}\right]=\frac{f_0-v_0}{f_0}\left[\frac{D}{f_e}\right]$

Distance between both the lenses,

$L=v_0+f_e$

Sign convention for solving numericals,

$u_0=-ve,v_0=+ve,f_0=+ve$

$u_e=-ve,v_e=-ve,f_e=+ve,m_0=-ve,m_e=+ve,M.P=-ve$

7.0Tube Length

• The distance l, i.e., the distance between the second focal point of the objective and the first focal point of the eyepiece $\left(focallength{f}_e\right)$ is called the tube length of the compound microscope.

Magnifying power of compound microscope when final image formed at infinity

$Tan\theta =\frac{h_0}{f_0}$

$Tan\theta =\frac{h_i}{l}$

$\frac{h_0}{f_0}=\frac{h_i}{l}\Rightarrow \frac{h_i}{h_0}=\frac{l}{f_0}$

So, M.P. of compound microscope when final image formed at ∞

$M.P=m_0\times m_e$

$M.P=-\frac{l}{f_0}\times \frac{D}{f_e}$

8.0Resolving Power of Microscope

• In the context of a microscope, the minimum distance between two lines at which they can still be distinguished is known as the Resolving Limit (RL). The reciprocal of this distance is referred to as the Resolving Power (RP).

$R.L.=\frac{\lambda }{2\mu \sin \theta }$

$R.P.=\frac{2\mu \sin \theta }{\lambda }$

$R.P.\propto \frac{1}{\lambda }$

$\lambda =\text{Wavelength of light used to illuminate the object,}$

$\mu =\text{The refractive index of the medium between the object and the objective lens.}$

$\theta =\text{The half-angle of the cone of light emitted from the point object,}$

$\mu \sin \theta =\text{Numerical Aperture}$

Note:An electron microscope uses an electron beam with a wavelength of approximately 1 Å, making its resolving power about 5,000 times greater than that of an ordinary microscope, which is around 5,000 Å.

9.0Difference Between a Simple Microscope And  Compound Microscope 

10.0Solved Examples On Microscopes

Q-1.In a compound microscope, the object is placed 6 cm from the objective lens with a focal length of 5 cm, and the focal length of the eyepiece is 10 cm. Calculate the magnifying power of the microscope when the final image is formed at near point.

Solution: When final image formed at least distance of distinct vision then,

$m_e=1+\frac{D}{f_e}=1+\frac{25}{10}=3.5$

$m_0=\frac{f_0}{f_0+u_0}=\frac{5}{5-6}=-5$

$M.P.=m_0\times m_e=-5\times 3.5=-17.5$

Q-2.An angular magnification of 30X is required using an objective with a focal length of 1.25 cm and an eyepiece(viewing lens) with a focal length of 5 cm. What distance should be set between the objective and the eyepiece of the compound microscope to form the final image at infinity?

Solution: If final image at ∞ then $m_e=\frac{D}{f_e}=\frac{25}{5}=5$

$M.P.=m_0\times m_e$

$-30=m_0\times 5\Rightarrow m_0=-6$

$m_0=\frac{f_0-V_0}{f_0}=-6\Rightarrow \frac{1.25-V_0}{1.25}=-6\Rightarrow V_0=8.75\text{ cm}$

Distance between objective and eyepiece is ,

$L=V_0+f_e=8.75+5=13.75\text{ cm}$

Q-3.A compound microscope has a magnifying power of 30X, with the focal length of its eyepiece being 5 cm. Assuming the final image is formed at a near point, calculate the magnification produced by the objective lens.

Solution:M.P. of eyepiece when final image formed at 25 cm away,

$m_e=1+\frac{D}{f_e}=1+\frac{25}{5}=6$

$M.P.=m_0\times m_e$

$-30=m_0\times 6=-5$

Q-4.The focal lengths of the objective & eyepiece lenses of a compound microscope are 1.2 cm and 3 cm, respectively. If the object is positioned 1.25 cm from the objective lens(Primary lens) and the final image is developed at infinity, determine the magnifying power of the microscope.

Solution:

$m_{\infty }=\frac{f_0}{(u_0-f_0)}\cdot \frac{D}{f_e}\Rightarrow m_{\infty }=-\frac{1.2}{(1.25-1.2)}\times \frac{25}{3}=-200$

Q-5.The objective and eyepiece lens of a microscope have focal lengths of 1 cm and 5 cm, respectively. Given that the magnifying power for a relaxed eye is 45, calculate the length of the microscope tube.

Solution: 

$m_{\infty }=\frac{(L_{\infty }-f_o-f_e)}{f_o{f}_e}\Rightarrow 45=\frac{(L_{\infty }-1-5)\times 25}{1\times 5}\Rightarrow L_{\infty }=15\text{ cm}$