Describe the construction of a compound microscope. Derive an expression for its total magnification. Draw a ray diagram for the formation of an image by a compound microscope.
Hint: A compound microscope is a powerful optical device that uses two convex lenses to magnify small objects. One lens (the objective) is positioned near the object, while the other lens (the eyepiece or ocular) is located closer to the viewer’s eye. The objective lens creates an enlarged, inverted image of the object, which then serves as an object for the eyepiece, providing additional magnification.
Formulas Used:
1. Lens Formula:
[latex]\frac{1}{f} = \frac{1}{v} – \frac{1}{u}[/latex]
Here, \( f \) is the focal length of the lens, [latex]v[/latex]is the image distance, and [latex]u[/latex] is the object distance.
2. Magnification of a Lens:
[latex]m = \frac{h’}{h} = \frac{v}{u}[/latex]
Where [latex]h[/latex] is the height of the object, [latex]h'[/latex] is the height of the image, [latex]u[/latex] is the object distance, and [latex]v[/latex] is the image distance.
3. Total Magnification:
[latex]m = m_o m_e[/latex]
Where [latex]m_o[/latex] is the magnification produced by the objective lens, and [latex]m_e[/latex] is the magnification produced by the eyepiece.
Step-by-Step Description of Construction:
A compound microscope is constructed using two convex lenses:
- Objective Lens:Placed close to the object being viewed. This lens has a short focal length and provides the primary magnification by forming a real, inverted, and magnified image of the object. This image is located between the objective and the eyepiece.
- Eyepiece Lens (Ocular): Positioned closer to the viewer’s eye, it has a longer focal length than the objective. The intermediate image produced by the objective lens acts as an object for the eyepiece, which further magnifies it to form a larger, virtual, and inverted image with respect to the original object.
Derivation of Total Magnification:
Magnification by the Objective Lens mo
If [latex]h[/latex] is the object height and [latex]h'[/latex] is the height of the intermediate image formed by the objective lens, we have:
[latex]m_o = \frac{h’}{h} = \frac{L}{f_o}[/latex]
where [latex]L[/latex] is the distance between the objective and eyepiece lenses, and [latex]f_o[/latex] is the focal length of the objective lens.
Magnification by the Eyepiece Lens me :
The magnification due to the eyepiece is given by:
me = [latex]\frac{D}{f_e} + 1[/latex]
where [latex]D[/latex] is the least distance of distinct vision (usually 25 cm), and [latex]f_e[/latex] is the focal length of the eyepiece.
Total Magnification (m):
[latex]m = m_o \times m_e = \left( \frac{L}{f_o} \right) \left( \frac{D}{f_e} + 1 \right)[/latex]
This expression gives the overall magnification of the compound microscope.
Ray Diagram for a Compound Microscope:
- Light from the object passes through the objective lens, forming an enlarged, inverted real image.
- This intermediate image then passes through the eyepiece, which magnifies it further.
- The final image viewed by the observer is virtual, larger, and inverted compared to the original object.
Note: The final magnification expression shows that to achieve a higher magnification, the focal lengths of both lenses should be minimized.