Q.A myopic adult has a far point at 0.1 m. His power of accomodation is 4 diopters.
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Microscope Magnification
A compound microscope views very small, nearby objects using two lenses in sequence: the objective (near the object) and the eyepiece (near the eye). Its total magnifying power is the product of what each lens contributes.
How the two lenses work together
- The object sits just beyond the focus of the short-focal-length objective (fo), which forms a real, inverted, magnified image inside the tube.
- That real image falls just inside the focus of the eyepiece (fe), which acts as a simple magnifier, producing a large virtual, magnified final image for the eye.
Because each stage magnifies, the effects multiply:
M=mo×me
The objective's magnification
mo=uovo≈foL
where L is the tube length (roughly the distance between the objective's second focal point and the eyepiece's first focal point). The object sits close to fo, so this approximation holds for a well-designed microscope.
The eyepiece's magnification
The eyepiece behaves as a simple magnifier:
- Final image at the near point (D=25 cm, largest magnification):
me=1+feD
- Final image at infinity (relaxed eye, "normal adjustment"):
me=feD
Total magnifying power
Image at the near point: M=foL(1+feD)
Image at infinity: M=foL⋅feD
High magnification needs short fo and fe (both sit in denominators) and a large tube length L — this is why a microscope objective is always a very short-focus lens.
Worked example
Objective fo=1.0 cm, eyepiece fe=2.5 cm, tube length L=20 cm, near point D=25 cm. Find M with the final image at the near point.
mo=1.020=20,me=1+2.525=11
M=20×11=220 …
Why this formula?
Microscope Magnification: Why the Formula Holds
Let's build the understanding from first principles — not just memorizing formulas, but seeing why they work.
1. What Does "Magnification" Mean in a Microscope?
A microscope creates a larger apparent image of a tiny object. The total magnification is the product of two stages:
- Objective lens — creates a real, enlarged, inverted image of the specimen.
- Eyepiece (ocular) — acts as a simple magnifier to view that real image.
So:
Total magnification = (magnification by objective) × (magnification by eyepiece)
2. The Key Formula
For a compound microscope in normal adjustment (final image at infinity, relaxed eye):
M=Mo×Me=(foL)×(feD)
Where:
- fo = focal length of objective
- fe = focal length of eyepiece
- L = tube length (distance between second focal point of objective and first focal point of eyepiece)
- D = near point distance of the eye (usually 25 cm)
3. Derivation of Objective Magnification Mo
Step 1: How the objective works
The objective lens forms a real, inverted, enlarged image of the specimen. The specimen is placed just outside its focal point (fo).
Step 2: Using the lens formula
For a thin lens:
vo1−uo1=fo1
(Using sign convention: uo is negative, vo is positive)
Step 3: The tube length approximation
In a standard microscope, the specimen is placed very close to fo, so:
- uo≈−fo (object just beyond focal point)
- The image is formed at the first focal point of the eyepiece, which is at a distance L from the second focal point of the objective.
Thus:
vo≈fo+L
Step 4: Magnification formula
Lateral magnification by objective:
Mo=∣uo∣vo≈fofo+L=1+foL
Since L≫fo in practice, 1 is negligible:
Mo≈foL
Why this makes sense: A shorter fo means the objective is more "powerful" — it bends light more sharply, creating a larger image at the fixed tube length.
4. Derivation of Eyepiece Magnification Me
Step 1: The eyepiece as a simple magnifier
The eyepiece takes the real image from the objective and acts like a magnifying glass. For relaxed eye (final image at infinity), the real image must be placed at the focal point of the eyepiece.
Step 2: Angular magnification
Angular magnification is defined as:
Me=angle subtended by object at near pointangle subtended by image at eye
For a simple magnifier with image at infinity:
Me=feD
Where D=25 cm (standard near point). …
Given: far point =0.1 m, accommodation =4 D, eye-lens-to-retina distance v=0.02 m.
(i) The correcting lens must image infinity at the far point: u=∞, v=−0.1 m.
P=v1−u1=−0.11−0=−10 D.
(ii) The relaxed eye focuses the far point on the retina (u=−0.1 m, v=+0.02 m):
Prelaxed=0.021−−0.11=50+10=60 D.
Full accommodation adds 4 D: Pmax=64 D. Near point un:
64=0.021−un1=50−un1⇒un=−141 m≈−0.071 m. …
A −10 D diverging lens corrects the distant vision; without glasses the near point lies at 141 m≈7.1 cm, and with the glasses it moves out to the normal 0.25 m (25 cm).
Data: far point =0.1 m; power of accommodation =4 D; image distance eye-lens to retina v=2 cm=0.02 m.
(i) Lens to see distant objects
A myopic eye focuses parallel light in front of the retina; the correcting lens must make an object at infinity appear at the far point. With u=∞ and v=−0.1 m (virtual, on the object side):
P=v1−u1=−0.11−0=−10 D.
So a diverging lens of power −10 D (focal length −10 cm) is required.
(ii) Near point without glasses
Treat the eye as a single lens forming the image on the retina, v=+0.02 m. When relaxed it is focused on the far point (u=−0.1 m):
Prelaxed=v1−u1=0.021−−0.11=50+10=60 D.
At maximum accommodation the eye adds 4 D:
Pmax=60+4=64 D.
The near point is the closest object still imaged on the retina:
Pmax=v1−un1⇒64=50−un1⇒un1=−14⇒un=−141 m≈−0.071 m.
So the unaided near point is about 7.1 cm from the eye.
(iii) Near point with glasses …
Method: Solving Eye-Defect (Myopia/Hypermetropia) Problems With the Lens Formula
Use this method for any question involving a defective eye's far point, near point, or power of accommodation, whether or not corrective spectacles are involved.
Steps
Step 1: Model the eye as a single lens imaging onto a fixed retina distance
Treat the eye's own lens as forming a real image at a fixed image distance v (the given eye-lens-to-retina distance), for whatever object distance u it is looking at. Every calculation below uses the thin-lens power formula
P=v1−u1
with all distances in metres so P comes out in dioptres.
Step 2: For the corrective-lens power, use the far point as the target image location
A corrective lens must take an object at infinity (u=∞) and form a virtual image exactly at the eye's own far point (myopia) — or bring a nearby object's image out to the near point (hypermetropia). Apply P=v1−u1 with v = the far/near point distance (negative if virtual, by sign convention) and u=∞ (so 1/u=0) to isolate P.
Step 3: For the eye's own near point (no glasses), add the accommodation power to the relaxed-eye power …
- GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL4 marksQ.A person with a normal near point (25 cm) using a compound microscope with objective of focal length 8.0 mm and an eyepiece of focal length 2.5 cm can bring an object placed at 9.0 mm from the objective in sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope.
›Reveal solutionSolution
First find where the objective lens forms its (real, magnified) intermediate image using the lens formula; then require the eyepiece to form the final image at the near point (25 cm) for maximum magnification, which fixes the eyepiece's object distance; the lens separation is the sum of the two image/object distances, and total magnification is the product of the two individual magnifications.
Given: near point D = 25 cm, objective focal length f_o = 8.0 mm = 0.8 cm, object distance from objective u_o = 9.0 mm = 0.9 cm (object just outside focal length, as required for a microscope objective), eyepiece focal length f_e = 2.5 cm.
Step 1 - Image formed by objective: Using the thin lens formula (sign convention: distances measured from the lens, object distance taken negative):
u_o = -0.9 cm, f_o = +0.8 cm (convex lens)
1/v_o - 1/u_o = 1/f_o
1/v_o = 1/f_o + 1/u_o = 1/0.8 + 1/(-0.9) = 1.25 - 1.111 = 0.139
v_o = 1/0.139 = 7.2 cm
So the objective forms a real, inverted, magnified image at 7.2 cm on the far side of the objective.
Step 2 - Eyepiece positioned for image at near point (maximum magnification, image at D): This intermediate image (from the objective) acts as the object for the eyepiece. For maximum magnifying power, the eyepiece should form its final (virtual) image at the near point, i.e. v_e = -25 cm (virtual image, same side as the object, by sign convention).
1/v_e - 1/u_e = 1/f_e
1/(-25) - 1/u_e = 1/2.5
-0.04 - 1/u_e = 0.4
-1/u_e = 0.44
u_e = -1/0.44 = -2.27 cm
So the intermediate image must be 2.27 cm in front of the eyepiece for the eyepiece to form the final image at the near point.
…
- GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL4 marksQ.Draw the ray diagram for the formation of image by a compound microscope and obtained the formula for magnification.
›Reveal solutionSolution
Figure — The stem instructs 'Draw the ray diagram for the formation of image by a compound microscope'; the catalog fig A compound microscope's objective forms a real, magnified image which then acts as the object for the eyepiece, which forms a further-magnified virtual image; total magnification is the product of the two individual magnifications.
Ray diagram (described): The object AB is placed just beyond the focus Fo of the objective lens (between fo and 2fo). The objective forms a real, inverted, magnified image A'B' beyond 2Fo on the other side. This intermediate image A'B' falls within the focal length of the eyepiece (between the eyepiece and its focus Fe). The eyepiece then acts as a simple magnifier on this intermediate image, producing a final virtual, further-magnified, and (relative to the object) inverted image A''B'' — formed at the near point D for maximum magnification, or at infinity for normal (relaxed-eye) adjustment. Both lenses are convex, mounted coaxially at opposite ends of a tube, with the objective having a small aperture and focal length, and the eyepiece having a larger aperture.
…
- GSEB Higher Secondary Certificate (HSC) Examination 2023Set ANNUAL4 marksQ.A compound microscope consists of an objective lens of focal length 2.0 cm and an eye piece of focal length 6.25 cm separated by a distance 15 cm. How far from the objective should an object be placed in order to obtain the final image ata) the least distance of distinct vision (25 cm) andb) at infinity? What is the magnifying power of microscope in each case?
›Reveal solutionSolution
Work back from the eyepiece: (a) for image at 25 cm, u_e = -5 cm, objective forms image at 10 cm, u_o = -2.5 cm, M = 20; (b) for image at infinity, u_e = -6.25 cm, objective image at 8.75 cm, u_o = -2.59 cm, M = 13.5.
Data: f_o = 2.0 cm, f_e = 6.25 cm, tube separation L = 15 cm, D = 25 cm.
- Final image at least distance of distinct vision (25 cm): Eyepiece: image virtual at v_e = -25 cm. 1/v_e - 1/u_e = 1/f_e: 1/(-25) - 1/u_e = 1/6.25 => -1/u_e = 1/6.25 + 1/25 = 5/25 = 1/5 => u_e = -5 cm. So the objective's image is 5 cm in front of the eyepiece; distance from objective = L - 5 = 10 cm => v_o = +10 cm. Objective: 1/v_o - 1/u_o = 1/f_o: 1/10 - 1/u_o = 1/2 => -1/u_o = 1/2 - 1/10 = 2/5 => u_o = -2.5 cm. Magnifying power: M = m_o (1 + D/f_e) = (v_o/|u_o|)(1 + D/f_e) = (10/2.5)(1 + 25/6.25) = 4 x 5 = 20.
- Final image at infinity: …
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