For a compound microscope under normal adjustment (final image at infinity), total magnification is:
\[ M = \left(\frac{L}{f_o}\right)\left(\frac{D}{f_e}\right) \]
Here \(L = 32\) cm, \(f_o = 2\) cm, \(f_e = 4\) cm and \(D = 25\) cm.
\[ M = \frac{32}{2} \times \frac{25}{4} \]
\[ M = 16 \times 6.25 = 100 \]
Correct Answer: 100
The compound microscope is one of the most important optical instruments discussed in the chapter Ray Optics. It is extensively tested in JEE Main and JEE Advanced because it integrates multiple fundamental ideas: refraction through lenses, magnification, angular size, focal length relationships and image formation principles. Understanding this topic deeply ensures clarity not only for direct numerical problems but also for conceptual and derivation-based questions.
A compound microscope consists of two convex lenses mounted coaxially. The lens closer to the object is called the objective lens. The lens through which the observer views is called the eyepiece. The objective has a very small focal length and small aperture. Its function is to produce a real, inverted and magnified intermediate image of the object. This image is formed within the tube of the microscope and serves as the object for the eyepiece.
The eyepiece behaves essentially like a simple magnifier. It magnifies the intermediate image produced by the objective. The final image formed by the microscope is virtual, inverted relative to the original object and highly magnified. The position of this final image depends on the adjustment of the microscope.
There are two standard adjustments considered in physics problems. The first is normal adjustment, where the final image is formed at infinity. The second is near point adjustment, where the final image is formed at the least distance of distinct vision, usually taken as 25 cm for a normal human eye.
Normal adjustment is preferred because when the final image is at infinity, the eye remains relaxed and does not need to exert accommodation effort. This reduces strain and is therefore practically useful and theoretically important in exam problems.
The total magnification of a compound microscope is the product of the magnification produced by the objective and that produced by the eyepiece. Mathematically,
\[ M = M_o \times M_e \]
The magnification of the objective is defined as the ratio of the image distance to the object distance. For small object distances close to the focal length, and under the usual approximations used in competitive exams, this magnification simplifies approximately to:
\[ M_o \approx \frac{L}{f_o} \]
Here L is the tube length of the microscope. The tube length is approximately the distance between the objective and the eyepiece. Increasing the tube length increases the objective magnification because the intermediate image is formed farther from the objective.
The magnification of the eyepiece under normal adjustment is the angular magnification of a simple magnifier when the final image is at infinity. This is given by:
\[ M_e = \frac{D}{f_e} \]
where D is the least distance of distinct vision and fe is the focal length of the eyepiece. Combining both contributions, the total magnification for normal adjustment becomes:
\[ M = \frac{L}{f_o}\times\frac{D}{f_e} \]
This formula is extremely important for JEE Main. Many direct questions are based on substitution in this expression. However, JEE Advanced may ask derivations or conceptual variations, so understanding the origin of the formula is essential.
It is important to note that magnification is dimensionless. Students sometimes mistakenly attach units such as “times” or cm, which is incorrect. Magnification simply represents a ratio of angular sizes.
If the final image is formed at the near point instead of infinity, the eyepiece magnification changes. In that case, the expression becomes:
\[ M = \frac{L}{f_o}\left(1+\frac{D}{f_e}\right) \]
This is a commonly tested variation. Therefore, students must carefully read whether the problem states “normal adjustment” or “final image at near point”.
Another important idea associated with microscopes is resolving power. Resolving power determines the ability of the microscope to distinguish two closely spaced objects. It depends on the wavelength of light used and the numerical aperture of the objective lens. Shorter wavelengths improve resolving power. This is why electron microscopes achieve much higher resolution than optical microscopes.
Microscope theory is closely related to telescope theory, but the two instruments serve different purposes. A telescope is used for viewing distant objects and its objective has large focal length. A microscope is used for viewing small nearby objects and its objective has very small focal length. Confusing the two formulas is a common mistake among students.
Magnification increases if the focal length of the objective decreases. This is why high-power microscopes use objectives with very small focal lengths. Similarly, decreasing the focal length of the eyepiece increases angular magnification. Increasing tube length also increases magnification, but practical limits exist due to instrument size and optical aberrations.
Common mistakes in exams include using telescope magnification formula instead of microscope formula, forgetting the value of D, using incorrect units and ignoring the type of adjustment mentioned in the question. Careful reading and strong conceptual clarity prevent such errors.
In JEE Main, questions from optical instruments are usually direct formula-based but require accuracy. In JEE Advanced, derivation-based, conceptual and multi-step problems may appear. Therefore, mastering the derivation once and understanding each term in the magnification expression is highly recommended.
Ultimately, the compound microscope is a beautiful application of geometric optics. It demonstrates how simple lens principles combine to create powerful instruments capable of observing microscopic structures. For competitive exams, remembering the correct formula and understanding the physical meaning behind it ensures confident and accurate problem solving.
This content is created strictly for educational and competitive examination preparation purposes only. It is not affiliated with NTA or any official examination authority.