By Fatskills Exam Guides Team — the exam nerds behind 28,500+ quizzes and 2.1M practice questions across 500+ global exams.
Hook: Mastering optical instruments unlocks 5-7 marks in IIT JEE (Main + Advanced) and helps you solve real-world problems like designing microscopes for medical research or telescopes for astronomy. One question on resolving power or magnification can be the difference between a 95% and 99%ile score.
Before diving in, ensure you understand: 1. Lens formula & sign conventions (real vs. virtual images, focal length signs). 2. Magnification (linear vs. angular, formula: m = v/u). 3. Ray optics basics (principal axis, focal point, ray diagrams).
If any of these are unclear, stop and review them first—this topic builds on them.
Problem: A simple microscope has a focal length of 5 cm. Calculate its angular magnification when the image is formed at the near point (D = 25 cm).
Solution: 1. Given: f = 5 cm, D = 25 cm. 2. Image is at near point → Use: [ M = 1 + \frac{D}{f} = 1 + \frac{25}{5} = 1 + 5 = 6 ] Answer: M = 6
What we did and why: We used the formula for angular magnification when the image is at the near point. The key was recognizing that D = 25 cm is a fixed value for the least distance of distinct vision.
Problem: A compound microscope has an objective of focal length 1 cm and an eyepiece of focal length 5 cm. The object is placed 1.1 cm from the objective. If the final image is formed at infinity, calculate: (a) Total magnification. (b) Length of the microscope.
Solution: 1. Objective lens: - Given: fo = 1 cm, uo = -1.1 cm (object distance is negative by sign convention). - Use lens formula: [ \frac{1}{f_o} = \frac{1}{v_o} - \frac{1}{u_o} \implies \frac{1}{1} = \frac{1}{v_o} - \frac{1}{-1.1} ] [ \frac{1}{v_o} = 1 - \frac{1}{1.1} = 1 - 0.909 = 0.091 \implies v_o = 11 \text{ cm} ] - Magnification of objective: [ M_o = \frac{v_o}{u_o} = \frac{11}{-1.1} = -10 \quad \text{(Negative sign indicates inversion)} ] 2. Eyepiece lens: - Final image is at infinity → Me = D/fe = 25/5 = 5. 3. Total magnification: [ M = M_o \times M_e = (-10) \times 5 = -50 \quad \text{(Negative sign indicates inversion)} ] 4. Length of microscope: [ L = v_o + f_e = 11 + 5 = 16 \text{ cm} ] Answers: (a) M = -50 (magnitude 50, inverted image) (b) L = 16 cm
What we did and why: We first found the image distance for the objective using the lens formula, then calculated its magnification. For the eyepiece, we used the formula for a relaxed eye (image at infinity). The total magnification is the product of the two, and the length is the sum of the image distance of the objective and the focal length of the eyepiece.
Problem: An astronomical telescope has an objective of focal length 100 cm and an eyepiece of focal length 5 cm. If the telescope is focused for a final image at infinity, calculate: (a) Angular magnification. (b) The distance between the objective and eyepiece. (c) If the diameter of the objective is 5 cm, calculate the resolving power for light of wavelength 500 nm.
Solution: 1. Angular magnification: [ M = \frac{f_o}{f_e} = \frac{100}{5} = 20 ] 2. Distance between lenses: [ L = f_o + f_e = 100 + 5 = 105 \text{ cm} ] 3. Resolving power: - Given: D = 5 cm = 0.05 m, λ = 500 nm = 500 × 10-9 m. - Formula: [ RP = \frac{D}{1.22 \lambda} = \frac{0.05}{1.22 \times 500 \times 10^{-9}} = \frac{0.05}{6.1 \times 10^{-7}} \approx 8.2 \times 10^4 ] Answers: (a) M = 20 (b) L = 105 cm (c) RP ≈ 8.2 × 104
What we did and why: For the telescope, we directly applied the formulas for angular magnification and length. For resolving power, we converted all units to meters and used the given wavelength. The key was unit consistency (cm → m, nm → m).
Mistake: Using D = 25 m instead of 25 cm for near point. Why it happens: Confusing units (meters vs. centimeters). Correct approach: Always use D = 25 cm for least distance of distinct vision.
Mistake: Ignoring sign conventions in lens formula. Why it happens: Forgetting that object distance (u) is negative for real objects. Correct approach: Use u = -ve for real objects, v = +ve for real images.
Mistake: Calculating total magnification as Mo + Me instead of Mo × Me. Why it happens: Confusing magnification with power (which is additive). Correct approach: Magnification is multiplicative for compound instruments.
Mistake: Using fo/fe for microscope magnification. Why it happens: Mixing up telescope and microscope formulas. Correct approach: For microscopes, use M = Mo × Me. For telescopes, use M = fo/fe.
Mistake: Forgetting to convert units in resolving power (e.g., cm → m, nm → m). Why it happens: Carelessness with unit consistency. Correct approach: Always convert to SI units (meters) before plugging into formulas.
Trap: Giving fo and fe in different units (e.g., fo = 1 m, fe = 5 cm). How to spot it: Units are mixed in the question. How to avoid it: Convert all units to the same system (e.g., cm or m) before calculations.
Trap: Asking for magnification when the final image is at the near point vs. infinity. How to spot it: The question specifies "image at D" or "image at infinity." How to avoid it: Use the correct formula for Me based on the final image position.
Trap: Providing μ (refractive index) for resolving power but not θ (half-angle). How to spot it: The question gives μ but not θ. How to avoid it: Assume θ is small and use sin θ ≈ θ (in radians) if not given, or look for NA = μ sin θ (numerical aperture).
"Listen up—this is your 60-second crash course for optical instruments in IIT JEE:
Now go crush that exam!"
Final Note for Teachers: - On camera: Use a whiteboard to draw ray diagrams for microscopes/telescopes while explaining. - Pacing: Spend 30% time on concepts, 50% on worked examples, 20% on common mistakes/traps. - Engagement: Ask students: "What’s the magnification if the final image is at infinity vs. near point?" to reinforce key differences.
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