Draw the geometry, declare the signs, then connect refraction and imaging to real optical instruments.
JEE Main · JEE Advanced · NEET UG85 min study guide11 indexed sections
BY THE END, YOU CAN
Apply reflection, Snell’s law, total internal reflection and prism geometry.
Use a consistent Cartesian convention for mirrors, lenses and spherical interfaces.
Analyse lens combinations and successive real/virtual objects.
Derive practical magnification relations while distinguishing resolution and aberrations.
BEFORE YOU START
Basic ray/wavefront ideas, refractive index and trigonometry.
Signed algebra, similar triangles and reciprocal relations.
Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.
CONNECT THE CONCEPT TO THE EXAM
Selected exam questions & worked solutions
These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.
Available in this lesson: JEE Main 2013; JEE Advanced 2021. This is a limited lesson sample, not a complete exam-question bank.
2 sourced sample questions available across the listed exams and years.
Matched topics: Reflection and Refraction.
JEE Main · 2013 · 2013-04-07 · Offline Paper 1 · Code P · Physics · printed Q27
Paraphrased exam question
Independently derived and checked solution. Archived key finality is unverified; this is not presented as an official final answer.
For rays transmitted through a triangular prism, select the deviation-versus-incidence graph: constant, smooth inverted-U, smooth U with a minimum, or a sharp V.
Reveal answer & worked solution
Answer: Smooth U with a minimum (independently derived; source option 3).
Deviation is δ=i+e−A, where A is prism angle and e is emergence angle.
The minimum occurs for symmetric passage: i=e and internal angles r₁=r₂=A/2.
On the transmitting branch, Snell’s law varies smoothly: deviation decreases toward this minimum and rises afterward, without a sharp corner.
Source: JEE (Main) 2013, 2013-04-07, Code P, Q27; original booklet scan archived by Resonance; independently derived solution; final-key status unverified. Solution written for this lesson.
Matched topics: Lens and Mirror Imaging.
JEE Advanced · 2021 · Paper 1 · Physics · Q3
Paraphrased exam question
Independently derived and checked solution. Archived key finality is unverified; this is not presented as an official final answer.
Use the intended thin-lens model in air. An object is 10 cm before a symmetric biconvex lens; a symmetric biconcave lens follows 10 cm later. Both have index 1.5 and surface-radius magnitudes 20 cm. Find total magnification.
Lensmaker’s formula gives focal lengths +20 cm and −20 cm.
With 1/v−1/u=1/f, the first lens has u₁=−10 cm, v₁=−20 cm and m₁=2.
The second lens sees u₂=−30 cm, yielding v₂=−12 cm and m₂=0.4. Hence m=m₁m₂=+0.8, an upright image.
Source: JEE (Advanced) 2021, Paper 1, Physics Q3; independently derived solution; key finality unverified. Solution written for this lesson.
01 / UNDERSTAND & APPLY
1. A ray diagram is a geometric model with a sign convention
Geometrical optics treats light as rays when relevant apertures and curvature scales are large compared with wavelength and diffraction can be neglected for the question. A ray indicates local propagation direction. A real image forms where rays physically converge; a virtual image is located where backward extensions appear to meet. A virtual image can be seen by an eye even though it cannot be caught directly on a screen at that apparent position.
Adopt one Cartesian convention for this lesson: light initially travels left to right, the pole or optical centre is the origin, rightward axial distances are positive and upward heights positive. Real objects on the incoming left side have u < 0. For lenses, a real image to the right has v > 0. For a mirror, a real reflected image in front on the left has v < 0 because the coordinate axis is still the original one.
A focal length is signed according to the focus location in this convention. A converging thin lens in air has f > 0 and a diverging lens f < 0. A concave mirror facing incoming light has f < 0 while a convex mirror has f > 0. These signs differ from some textbooks’ positive-real conventions; never combine equations from one convention with signs from another.
Paraxial formulas assume rays close to the principal axis and small angles so sinθ ≈ tanθ ≈ θ in radians. Spherical aberration means rays far from the axis may not share one image point. A clean schematic ray diagram illustrates the model, but it does not remove the approximation behind the compact imaging equation.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
Quantitative paraxial geometry. Solid segments are actual light paths; dashed segments are mathematical extensions, not rays travelling backward.
02 / UNDERSTAND & APPLY
2. Reflection and refraction depend on the interface normal
The law of reflection gives equal incident and reflected angles measured from the local normal. The incident ray, reflected ray and normal lie in one plane. For a plane mirror, the image is as far behind the mirror as the object is in front, has the same size and is virtual for an ordinary real object. Apparent left–right reversal is best understood as reversal of the coordinate perpendicular to the mirror rather than a mysterious exchange of every horizontal direction.
At a stationary interface between isotropic transparent media, Snell’s law is n₁ sin i = n₂ sin r. Angles are measured from the normal, not the surface. A ray entering higher refractive index bends toward the normal unless it is normally incident. Frequency remains fixed; speed and wavelength decrease when index increases.
For a plane-parallel slab surrounded by the same medium on both sides, the emergent ray is parallel to the incident ray but laterally displaced. For slab thickness t, incidence i and refraction r, displacement magnitude is t sin(i − r)/cos r. This result assumes flat parallel faces and the stated surrounding media. A prism generally changes the final direction because its faces are not parallel.
For near-normal viewing through a plane boundary from index n₁ toward a viewing medium n₂, an object at real depth d in n₁ appears at depth approximately d n₂/n₁. The approximation becomes angle-dependent away from the paraxial direction. Distinguish apparent depth from lateral displacement and from optical path length.
ORIGINAL WORKED EXAMPLE
Refraction and apparent depth
A ray travels from air into glass of index 1.5 with sin i = 0.75. Find sin r. A mark 12 cm beneath a flat water surface of index 4/3 is viewed nearly normally from air; find apparent depth.
Snell’s law gives sin r = sin i/1.5 = 0.50, so r = 30°.
For the water mark, apparent depth is real depth divided by water’s index relative to air: 12/(4/3) = 9 cm.
The second result is a near-normal imaging approximation and is not found by treating the mark as a ray angle.
Answer r = 30°; apparent depth 9 cm.
03 / UNDERSTAND & APPLY
3. Total internal reflection and prism deviation have different conditions
Total internal reflection requires light to travel from higher index n₁ toward lower index n₂ and incidence above the critical angle ic, where sin ic = n₂/n₁. At the critical angle, the refracted ray grazes the interface. Below it, some light transmits and some reflects; above it, the ideal lossless interface has no propagating refracted ray in the second medium.
Optical fibres guide light using an appropriate higher-index core and lower-index cladding. The ray model uses repeated total internal reflection, while a full waveguide description is needed for detailed modes. A prism can also use total internal reflection to redirect a beam efficiently. “Large incidence angle” alone is insufficient without the high-to-low index condition.
For a prism of apex angle A, internal refraction angles satisfy r₁ + r₂ = A, and deviation is δ = i + e − A. At minimum deviation in a symmetric path, i = e and r₁ = r₂ = A/2. For a prism in air, its index is n = sin[(A + δmin)/2]/sin(A/2). The thin-prism approximation δ ≈ (n − 1)A uses small angles in radians.
Dispersion arises because refractive index depends on wavelength. In common normal-dispersion glass, shorter visible wavelengths typically have greater index and deviation, but the detailed material dispersion matters. A prism changes the propagation directions of colours without changing their frequencies at stationary boundaries.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
Direction schematics, not numerical angle diagrams. TIR requires higher-to-lower refractive index and incidence above the critical angle.
ORIGINAL WORKED EXAMPLE
A critical-angle test
Light in glass of index 1.5 meets an air boundary at 50°. Does total internal reflection occur?
Critical angle satisfies sin ic = 1/1.5 = 2/3, giving ic ≈ 41.8°.
The ray travels high index to low and 50° exceeds 41.8°.
Both necessary conditions are met.
Answer Yes, in the ideal lossless interface model.
04 / UNDERSTAND & APPLY
4. Spherical mirrors and thin lenses use different signed equations
For a spherical mirror in the stated Cartesian convention, 1/v + 1/u = 1/f with f = R/2, and transverse magnification m = himage/hobject = −v/u. The radius R is signed by the centre of curvature’s location. A concave mirror can make real inverted or virtual upright images depending on object position; a convex mirror gives a reduced upright virtual image for an ordinary real object.
For a thin lens with the same surrounding medium on both sides, 1/v − 1/u = 1/f and m = v/u. A converging lens with an object beyond its focal distance forms a real inverted image. With the object between the lens and its focus, it forms a virtual upright magnified image. A diverging lens forms a reduced upright virtual image for a real object in the usual setup.
Three useful paraxial lens rays are: a ray parallel to the axis refracts through the far focus for a converging lens; a ray through the optical centre is approximately undeviated; and a ray directed through the near focus emerges parallel. For a diverging lens, use backward extensions of the appropriate refracted rays. At least two rays locate the image, and a third can check the construction.
For mirrors, a parallel incident ray reflects through the focus for a concave mirror; a ray through the centre of curvature retraces its path; and a ray through the focus reflects parallel. Keep actual ray segments and backward extensions visually distinct. A dashed construction behind a mirror or on the object side of a lens is not an actual light path.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
Quantitative paraxial ray construction with a common spatial scale. Actual rays are solid and travel left to right; Cartesian u<0,v>0,f>0.
ORIGINAL WORKED EXAMPLE
A converging lens with a real image
An object is 30 cm left of a thin converging lens of focal length 20 cm in air. Find image position and magnification using the stated Cartesian convention.
Use u = −30 cm and f = +20 cm in 1/v − 1/u = 1/f.
1/v + 1/30 = 1/20, so 1/v = 1/60 and v = +60 cm.
Magnification is v/u = 60/(−30) = −2, meaning a real inverted image twice the object height.
Answer 60 cm right of lens; magnification −2.
ORIGINAL WORKED EXAMPLE
A concave mirror with the same axis convention
An object is 30 cm in front of a concave mirror of focal length magnitude 10 cm. Find image position and magnification.
With incoming light left to right, u = −30 cm and the concave focus lies left, so f = −10 cm.
1/v + 1/(−30) = 1/(−10), giving 1/v = −1/15 and v = −15 cm.
m = −v/u = −(−15)/(−30) = −0.5. The image is real, inverted and half-sized in front of the mirror.
Answer 15 cm in front; magnification −0.5.
05 / UNDERSTAND & APPLY
5. Curvature and refractive index set lens power
For refraction at a spherical interface from index n₁ to n₂ in the Cartesian convention, n₂/v − n₁/u = (n₂ − n₁)/R. The centre of curvature to the right gives R > 0. This paraxial relation can be applied successively to the two surfaces of a lens, tracking the first image as the second surface’s object.
For a thin lens of refractive index nlens in a surrounding medium nmedium on both sides, the lens-maker relation is 1/f = (nlens/nmedium − 1)(1/R₁ − 1/R₂). Surface radii must be signed with the same global axis. A symmetric biconvex lens in air has R₁ > 0 and R₂ < 0, giving positive power when nlens > nmedium.
A lens’s shape alone does not decide convergence independently of the surrounding medium. If its index equals the medium’s, its refractive power vanishes in the ideal model. If immersed in a medium with larger index, the sign can reverse. This is a useful limiting-case test of the relative-index factor.
Optical power is P = 1/f with f in metres, measured in dioptres. Thin lenses in contact have powers that add: Ptotal = P₁ + P₂, provided the ordinary thin-lens same-medium approximation applies. Separated lenses require propagation between them; simply adding powers then generally misses the separation term.
ORIGINAL WORKED EXAMPLE
Two lenses in contact
A +10 cm converging lens is in contact with a −20 cm diverging lens in air. Find equivalent focal length and power.
Convert to metres: powers are +1/0.10 = +10 D and 1/(−0.20) = −5 D.
Total power is +5 D, so equivalent focal length is 1/5 m = 0.20 m.
The combination remains converging, but less strongly than the first lens alone.
Answer +20 cm; +5 D.
06 / UNDERSTAND & APPLY
6. A first image becomes the next element’s object
For separated optical elements, solve one at a time. Locate the first image in a common coordinate system, then measure its distance from the second element. If rays arriving at the second element are already converging toward a point beyond it, that point is a virtual object for the second element, with positive u in the stated left-to-right convention.
An intermediate image need not be physically formed before the second element intercepts the rays. The would-be convergence point still determines the incoming wavefront curvature and serves as the second element’s object. This is why negative or positive object distances cannot be chosen solely from the word “image” in the previous step.
Total transverse magnification is the product of the successive signed magnifications when image heights are consistently referenced. A negative product indicates overall inversion. Distances must be recomputed after every reflection if the convenient propagation-axis convention changes; either maintain a global Cartesian axis carefully or explicitly restart the convention at each stage.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
Paraxial coordinate construction. Horizontal scale10 pixels/cm; vertical scale is exaggerated. Dashed lines are would-be paths with lens2 absent, not actual transmitted paths. The final image is real.
ORIGINAL WORKED EXAMPLE
A virtual object for the second lens
An object lies 30 cm left of a +10 cm thin lens. A second +20 cm lens is 5 cm to its right. Find the final image relative to the second lens.
For lens 1, 1/v₁ + 1/30 = 1/10, giving v₁ = 15 cm right of lens 1.
That point lies 10 cm to the right of lens 2. Rays reaching lens 2 are converging toward it, so u₂ = +10 cm, a virtual object.
For lens 2, 1/v₂ − 1/10 = 1/20, so 1/v₂ = 3/20 and v₂ = 20/3 cm.
The final image is real and about 6.67 cm to the right of lens 2.
Answer 20/3 cm to the right of the second lens.
07 / UNDERSTAND & APPLY
7. Optical instruments magnify angle, not just image height
The eye responds to the angular size of an object’s image on the retina. A simple magnifier lets an object be placed closer than the unaided near-point distance D while forming a virtual image the eye can focus on. For relaxed-eye viewing with the object near the magnifier’s focal plane, angular magnification is approximately D/f. With the final virtual image at the near point, it is approximately 1 + D/f under the usual close-lens geometry.
A compound microscope uses a short-focal-length objective to form a real enlarged intermediate image, then an eyepiece as a magnifier. For normal adjustment and a suitable long-tube paraxial approximation, magnification magnitude is approximately (L/fobjective)(D/feyepiece), where L is the relevant optical tube-length approximation. Exact problems should use the objective’s actual image/object distances rather than treating any physical tube measurement as interchangeable with L.
An astronomical refracting telescope for a distant object uses the objective to form an image near its focal plane and the eyepiece to make the final image at infinity in normal adjustment. Angular magnification is −fo/fe for two converging lenses, with the minus indicating inversion. Lens separation is approximately fo + fe. The objective aperture controls light collection and diffraction-limited resolution, not only magnification.
For basic vision correction, myopia requires a diverging lens to make a distant object appear at the eye’s far point, while hypermetropia/near-vision difficulty may require a converging lens to place a desired near object’s virtual image at the eye’s accessible near point. These are idealised optical exercises, not a prescription for an actual person; eye examination also considers other optical and health factors.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
Paraxial construction with horizontal scale10 pixels/cm and vertical scale100 pixels/cm, exaggerated to separate rays. Drawn angles are not measured angles. No final finite-distance screen image is formed.
ORIGINAL WORKED EXAMPLE
A telescope in normal adjustment
A refracting telescope has objective focal length 100 cm and eyepiece focal length 5 cm, both converging. Find angular magnification and separation for a distant object with relaxed-eye viewing.
Normal adjustment places the objective’s real focal-plane image at the eyepiece’s front focal plane.
Angular magnification is −fo/fe = −100/5 = −20.
Lens separation is fo + fe = 105 cm. The negative sign means the final view is inverted relative to the object.
Answer Magnification −20; separation 105 cm.
08 / UNDERSTAND & APPLY
7a. A microscope needs an actual intermediate-image calculation
For a specified microscope geometry, calculate the objective image distance before using any tube-length approximation. In normal adjustment the real intermediate image must lie one eyepiece focal length before the eyepiece, so the physical lens separation is v_objective + f_eyepiece. The objective magnification is v_objective/u_objective in the Cartesian convention. Multiplying it by D/f_eyepiece gives the signed angular magnification for a relaxed eye close to the eyepiece.
The familiar L/f_objective approximation replaces the exact objective image/object distance ratio using a long-tube regime. A given mechanical tube length may include distances that differ from the optical quantity used in that approximation. An explicit separation and exact thin-lens calculation prevent that ambiguity. The intermediate image is real and inverted; the eyepiece supplies angular enlargement rather than turning that image upright.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
Exact thin-lens distances for the worked model. Horizontal scale200/11 pixels/cm; vertical exaggeration separates rays, so drawn ray angles are not measured angles. The tiny object arrow is explicitly labelled.
ORIGINAL WORKED EXAMPLE
A compound microscope without a tube-length shortcut
A thin objective has focal length 1.0 cm and an object 1.1 cm to its left. A thin eyepiece has focal length 2.5 cm. Choose separation for a final image at infinity and find angular magnification using reference near point D = 25 cm. Treat the supplied lengths as model values.
For the objective, 1/vₒ − 1/(−1.1) = 1/1.0, giving 1/vₒ = 1 − 10/11 = 1/11 cm⁻¹. Thus vₒ = 11 cm.
The intermediate image must be 2.5 cm left of the eyepiece, so lens separation is 11 + 2.5 = 13.5 cm.
Objective transverse magnification is 11/(−1.1) = −10. The relaxed-eye eyepiece angular factor is 25/2.5 = 10.
Total angular magnification is −10 × 10 = −100. This calculation used the actual objective distances; it did not identify 13.5 cm with a long-tube L.
Answer Lens separation 13.5 cm; signed angular magnification −100 in the stated paraxial model.
09 / UNDERSTAND & APPLY
7b. Reflection sends the real image back toward the object side
A concave mirror with an ordinary real object beyond the centre of curvature forms a smaller real image between the focus and centre of curvature. Both object and real image lie in front of the mirror, even though the incident and reflected rays travel in opposite directions. In the global left-to-right Cartesian convention used here, those object-side coordinates are negative.
The diagram uses an incident ray parallel to the principal axis, which reflects through the focus, and a ray aimed through the centre of curvature, which retraces its incoming line. Their reflected segments meet at the inverted real image. This is qualitatively different from transmission through a lens: a real mirror image does not have to lie to the right of the optical element. The paraxial construction represents a spherical mirror; wide-angle rays do not all meet at one exact point.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
Paraxial pole-plane construction for a concave spherical mirror, not a claim that a flat mirror focuses. Horizontal scale10 pixels/cm; vertical exaggeration makes image heights visible. P is the pole, F focus and C centre of curvature.
10 / UNDERSTAND & APPLY
8. Magnification, brightness and resolution are not synonyms
A large image is not necessarily a detailed image. Diffraction limits the smallest angular separation an aperture can resolve; geometric aberrations can blur it further. Increasing an eyepiece’s magnification beyond the available information merely enlarges blur. Lens diameter, optical quality and wavelength matter alongside focal length.
Spherical aberration arises when rays far from the principal axis focus differently from paraxial rays at a spherical surface. Chromatic aberration arises because refractive index and thus focal length vary with wavelength. Aperture stops can reduce some geometric aberrations at the cost of light collection, while compound lens designs can correct selected aberrations over a useful range.
Ray diagrams should therefore label which rays are actual and which are extensions, and whether the model is thin, paraxial and monochromatic. A mathematically divergent image distance at u = −f for the stated lens convention means emerging rays are parallel in the ideal model, not that energy or image brightness becomes infinite.
Use limiting cases to check signs: a distant real object should focus near the image-side focus of a converging lens; an index-matched lens should have zero power; a plane mirror should preserve size; and a diverging lens with a real object should not suddenly produce a real image in the same simple setup.
11 / UNDERSTAND & APPLY
9. Original practice: geometry before substitution
Use the Cartesian convention stated in section 1. Distances are measured from the relevant optical centre or mirror pole; all lenses are thin and rays paraxial unless stated otherwise.
Angles of incidence/refraction are measured from normals.
Total internal reflection needs high-to-low index and incidence above critical.
Mirror and lens equations have different signs in the same Cartesian convention.
An intermediate image can become a virtual object for the next element.
Instrument magnification is angular and has viewing-condition assumptions.
Ray optics does not remove diffraction or aberration limits.
ORIGINAL PRACTICE · 1
A real object is 10 cm before a plane mirror. Where is its image and what is magnification?
Reveal answer & reasoning
10 cm behind; +1.
The image is virtual, upright and equal in size at the symmetric position.
ORIGINAL PRACTICE · 2
Can total internal reflection occur from air into ordinary glass?
Reveal answer & reasoning
No.
It requires travel from higher to lower refractive index.
ORIGINAL PRACTICE · 3
A medium has critical angle 30° to air. Find its index.
Reveal answer & reasoning
2.
sin ic = 1/n gives n=1/sin30°.
ORIGINAL PRACTICE · 4
A +15 cm lens has an object 30 cm to its left. Find v and m.
Reveal answer & reasoning
v = +30 cm; m = −1.
1/v +1/30=1/15, so v=30; m=v/u=−1.
ORIGINAL PRACTICE · 5
A −10 cm lens has a real object 20 cm left. Find image position.
Reveal answer & reasoning
v = −20/3 cm.
1/v +1/20=−1/10, so 1/v=−3/20. The image is virtual on the left.
ORIGINAL PRACTICE · 6
An object is 5 cm left of a +10 cm lens. Find v and m.
Reveal answer & reasoning
v = −10 cm; m = +2.
1/v+1/5=1/10 gives v=−10. The virtual image is upright and enlarged.
ORIGINAL PRACTICE · 7
A concave mirror has R=−40 cm. Find focal length in this convention.
Reveal answer & reasoning
−20 cm.
For paraxial spherical-mirror imaging f=R/2.
ORIGINAL PRACTICE · 8
Two contact lenses have powers +4 D and −1 D. Find total power and focal length.
Reveal answer & reasoning
+3 D; +1/3 m.
Contact powers add, then f=1/P in metres.
ORIGINAL PRACTICE · 9
A glass lens is immersed in material of the same refractive index. What is its ideal refractive power?
Reveal answer & reasoning
Zero.
The relative-index factor nlens/nmedium−1 vanishes.
ORIGINAL PRACTICE · 10
A magnifier has f=5 cm and the conventional near distance D=25 cm. Find relaxed-eye angular magnification.
Reveal answer & reasoning
5.
M≈D/f=25/5.
ORIGINAL PRACTICE · 11
A telescope’s objective focal length doubles with eyepiece fixed. What happens to normal-adjustment angular magnification magnitude?
Reveal answer & reasoning
It doubles.
|M|=fo/fe, though other performance measures do not follow from magnification alone.
ORIGINAL PRACTICE · 12
Why must a second lens sometimes use a positive object distance?
Reveal answer & reasoning
The incoming rays may already be converging toward a point beyond it.
That would-be point is a virtual object for the second lens in the stated Cartesian convention.
KEEP THE BIG IDEAS
Quick revision
A sign convention is part of the model, not an afterthought.
Actual rays locate real images; extensions locate virtual images.
Curvature and relative refractive index determine power.
Optical instruments trade angular size, light collection and resolution under explicit assumptions.
The teaching notes, worked explanations and original practice are written for this lesson. Separately labeled selected exam questions identify their exam, year, source and whether wording is paraphrased or an exact excerpt. Educational references do not certify current exam coverage.