Optical Instruments & Magnification
Microscope
Explore the optics of a compound microscope. Interact with a school microscope model, adjust focus knobs to resolve biological specimens, switch objective turrets, and look through the eyepiece to see specimens magnified.
Compound Microscope Laboratory
Turn the focus knobs and adjust illumination to resolve the cell structures inside the viewport.
Microscope Telemetry
Out of Focus- Specimen slide
- Onion Cells
- Objective Magnification
- 10 x
- Eyepiece Magnification
- 10 x
- Total Magnification
- 100 x
- Field of View (diameter)
- 1.80 mm
Working Principle of a Compound Microscope
A compound microscope is a classic optical instrument that uses a multi-lens structure to produce highly magnified virtual images of small near objects.
- Objective Lens: A short focal length convex lens located close to the specimen. It bends light to form an intermediate real, inverted, and magnified image inside the body tube.
- Eyepiece (Ocular): A convex lens of moderate focal length. It acts as a simple magnifier, enlarging the intermediate image to create a virtual, magnified final image for the observer\'s eye.
- Total Magnification: The total magnifying power is the product of both lens systems:
Magnification Formula
Mtotal = Mobj × Meye
Resolution Limits
While magnification makes things larger, resolution determines structural clarity. Microscope resolving power depends on light wavelength and the lens numerical aperture (NA).
Tube Length Formula
For a microscope with tube length L, near point D, and objective/eyepiece focal lengths fo, fe:
Approximate Magnification
M ≈ - (L / fo) × (D / fe)
Focusing Actions
Turning the coarse knob raises or lowers the stage quickly to align the specimen inside the objective\'s shallow focal field. The fine knob allows micrometer tweaks to make cell boundaries look sharp.
Step-by-Step Solved Problems
Master magnification math and microscope optics with these step-by-step solutions.
Example 1 Problem Statement
A compound microscope consists of an objective lens of focal length 1.0 cm and an eyepiece of focal length 5.0 cm. If the distance between the two lenses (tube length L) is 20 cm, calculate the total magnification of the microscope when the final image is formed at the near point (25 cm).
View Mathematical Solution Steps
- Given parameters: focal length of objective fo = 1.0 cm, focal length of eyepiece fe = 5.0 cm, tube length L = 20 cm, near point distance D = 25 cm.
- Magnification of the objective: mo ≈ -L / fo = -20 / 1.0 = -20.
- Magnification of the eyepiece (acting as a simple magnifier focusing at near point): me = 1 + D/fe = 1 + 25/5 = 1 + 5 = 6.
- Total magnification is the product: m = mo × me = -20 × 6 = -120.
- The negative sign indicates the final image is inverted.
Final Derived Answer: Total Magnification M = 120x (inverted).
Example 2 Problem Statement
The objective of a compound microscope has a magnification of 10x, and its eyepiece has a magnification of 15x. If the field of view under this setup is measured to be 1.8 mm, find the total magnification and estimate the field of view when switching to a 40x objective with the same eyepiece.
View Mathematical Solution Steps
- Step 1: Find initial total magnification: Minitial = 10 × 15 = 150x.
- Step 2: Find final total magnification with 40x objective: Mfinal = 40 × 15 = 600x.
- Step 3: Recall the relation between field of view (FOV) and magnification: FOVfinal / FOVinitial = Minitial / Mfinal.
- Step 4: Solve for final FOV: FOVfinal = 1.8 mm × (150 / 600) = 1.8 × 0.25 = 0.45 mm.
Final Derived Answer: Total Magnification = 600x, Field of View = 0.45 mm (450 microns).
Example 3 Problem Statement
A microscopic specimen is placed 1.2 cm in front of an objective lens of focal length 1.0 cm. Find the position of the intermediate real image formed by the objective lens.
View Mathematical Solution Steps
- Identify object distance for objective: uo = -1.2 cm.
- Identify focal length of objective: fo = +1.0 cm (convex lens).
- Apply the lens formula: 1/vo - 1/uo = 1/fo.
- Substitute values: 1/vo - 1/(-1.2) = 1/1.0.
- Solve for 1/vo: 1/vo = 1.0 - 0.833 = 0.167 cm⁻¹.
- Invert to find image distance: vo = 1 / 0.167 = +6.0 cm.
Final Derived Answer: Intermediate Real Image Position vo = 6.0 cm behind the objective lens.
Self-Check Questions
Question 1
Identify the primary optical difference between the objective lens and the eyepiece lens in a compound microscope.
Show Answer & Explanation
The objective lens has a very short focal length and small aperture to capture light from a near specimen and form a highly magnified intermediate real image. The eyepiece has a longer focal length and larger aperture to magnify this intermediate image and act as a comfortable viewport for the eye.
Question 2
Why is the intermediate image formed by the objective lens real, while the final image seen by the observer is virtual?
Show Answer & Explanation
The specimen is placed just outside the focal point of the objective lens, producing a real, inverted, and magnified intermediate image. This intermediate image is designed to fall inside the focal length of the eyepiece lens, which then acts as a simple magnifying glass to produce a further enlarged, virtual final image.
Question 3
What is the function of the sub-stage condenser on a school microscope?
Show Answer & Explanation
The condenser is a lens system under the stage that gathers light from the lamp or mirror and concentrates it into a tight cone directly onto the specimen, optimizing contrast, brightness, and resolution.
Question 4
How does field of view change when switching from a low-power to a high-power objective lens?
Show Answer & Explanation
As magnification increases, the field of view (the physical diameter of the circular area visible through the eyepiece) decreases proportionally, meaning you see a smaller section of the specimen in greater detail.
Question 5
What is resolving power of a microscope, and how does it differ from magnifying power?
Show Answer & Explanation
Magnifying power is the ability to make an object appear larger. Resolving power is the ability of the microscope to distinguish between two close adjacent points as separate structures, defining the clarity and level of fine detail in the image.
Question 6
Explain why a cover slip is placed over a wet mount biological specimen on a microscope slide.
Show Answer & Explanation
A cover slip is used to flatten liquid specimens to a uniform thickness, protect the microscope's objective lens from contacting the liquid, and keep the specimen in a single focal plane.