CLASS 12 PHYSICS · CHAPTER 9 · STUDY NOTES
Ray Optics and Optical Instruments
A comprehensive study of light as rays, exploring reflection by spherical mirrors, refraction through various media, total internal reflection, and the working principles of optical instruments like microscopes and telescopes.
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Introduction to Ray Optics
Choose ray optics when the scale of the opening or object is large compared with the wavelength.A ray shows a direction, not a thin physical thread
Read the assumptions and limits
Refractive index describes optical speed, not mass density. Frequency stays unchanged at a stationary boundary. The simple apparent-depth relation assumes near-normal viewing. Diffraction is treated separately in Wave Optics.
Original teaching diagrams · NCERT Physics Part II, chapter 9, PDF pages 1–2 · learner-controlled traces
- Ray
- A line indicating the local direction of propagation. In an isotropic uniform medium it is normal to a wavefront.
A wavefront joins points at the same phase. Rays perpendicular to that front show the propagation direction; they are a representation, not narrow physical strings of light.
Straight rays in a uniform medium let us construct images. At boundaries they reflect or refract. A small opening can produce significant diffraction, so a purely geometric description has limits.
Let’s look a little closer.
Go deeper: When the model is useful
Visible light has a wavelength of a few hundred nanometres. A centimetre-wide lens is many wavelengths across, making geometric rays useful for its main imaging behaviour.
Even large optical instruments have diffraction and aberrations. A ray diagram predicts the geometric image, rather than the entire brightness distribution at that image.
Take a moment. What do you think?
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Question
A 500 nm wave passes through an opening 5 mm wide. How many wavelengths fit across the opening, and is ray optics a useful first approximation?
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Reflection of Light by Spherical Mirrors
Find the normal before measuring a reflection angle.The local normal tells a reflected ray where to go
Read the assumptions and limits
Thin lenses and small angles near the axis are assumed for image equations. Drawings are teaching schematics: instruments can use different horizontal and vertical scales. Solid paths show light travelling; dashed paths locate virtual images. Real devices have finite apertures and aberrations. D = 25 cm is a conventional reference near point, not a universal measurement for every eye.
Original teaching diagrams · NCERT Physics Part II, chapter 9, PDF pages 2–4 · learner-controlled traces
- Principal focus
- For paraxial parallel rays, the point where reflected rays meet, or from which they appear to diverge.
The laws of reflection use the normal at the point where light strikes: i = r, and incident ray, reflected ray and normal lie in one plane. At a spherical surface, that normal passes through the centre of curvature C.
A concave mirror converges paraxial parallel rays to a real focus; a convex mirror gives a virtual focus behind it. For a small aperture near the axis, f ≈ R/2 with signed Cartesian distances.
Go deeper: Why “paraxial” matters
Rays close to the axis make small angles. Approximations such as sin θ ≈ tan θ ≈ θ let a spherical mirror have one focal length in this model.
Rays far from the axis of a spherical mirror need not meet at that same point: this is spherical aberration. Parallel rays at a small angle to the axis meet near the focal plane, away from the principal focus.
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Question
A concave spherical mirror has radius magnitude 48 cm. Estimate its focal length using the Cartesian convention for light incident from the left.
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The Mirror Equation and Magnification
Follow both the real and virtual branches as the object approaches the focus.Move the object — follow the image
Read the assumptions and limits
Thin lenses and small angles near the axis are assumed for image equations. Drawings are teaching schematics: instruments can use different horizontal and vertical scales. Solid paths show light travelling; dashed paths locate virtual images. Real devices have finite apertures and aberrations. D = 25 cm is a conventional reference near point, not a universal measurement for every eye.
Original teaching diagrams · NCERT Physics Part II, chapter 9, PDF pages 4–6 · learner-controlled traces
- Mirror equation
- 1/v + 1/u = 1/f, with distances measured from the pole under the Cartesian sign convention; transverse magnification m = −v/u.
Take the incident direction as positive x. A real object on the left has u < 0; a concave mirror has f < 0. A real reflected image is also on the left, so v < 0. A virtual image behind the mirror has v > 0.
Magnification compares signed heights. A negative m means an inverted image; a positive m means an erect image. At u = f there is no finite image distance: reflected rays are parallel.
Go deeper: The graph uses a stated alternative coordinate
The interactive graph uses positive physical object distance d and positive real reflected distance v_real. Its equation is v_real = fd/(d−f), with f here the positive concave focal-distance magnitude. Do not insert those graph coordinates directly into the Cartesian equation.
Construct rays first, then use the formula to check them. Dashed extensions locate a virtual image; light does not travel backwards along the extensions.
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Question
A concave mirror has f = −15 cm and a 2 cm object at u = −45 cm. Find v and the image height.
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Refraction of Light
Separate what changes at the boundary from what remains the same.The boundary changes direction and wavelength
Read the assumptions and limits
Refractive index describes optical speed, not mass density. Frequency stays unchanged at a stationary boundary. The simple apparent-depth relation assumes near-normal viewing. Diffraction is treated separately in Wave Optics.
Original teaching diagrams · NCERT Physics Part II, chapter 9, PDF pages 8–9 · learner-controlled traces
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- Refractive index
- Absolute index n = c/v_phase. At a stationary interface, n₁ sin i = n₂ sin r; angles are measured from the normal.
At an air-to-glass boundary, light normally travels more slowly and bends towards the normal. The oscillation frequency is fixed by the source and remains unchanged across a stationary boundary; wavelength changes because v = fλ.
A parallel slab has two refractions. With the same external medium on both sides, the emergent ray is parallel to the incident ray but shifted sideways. The shift depends on thickness and incidence.
Go deeper: Optical density and apparent depth
A larger refractive index describes a lower phase speed, not necessarily a greater mass density. Relative index n₂/n₁ is the ratio needed for a particular pair of media.
An object under water appears shallower because the emerging rays diverge more and their backward extensions meet above it. Apparent depth ≈ real depth/n for near-normal viewing from air. At oblique angles the simple ratio is not generally exact.
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Question
A 600 nm wave in air enters glass of index 1.50 at 30°. Find the refracted angle and wavelength in glass.
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Total Internal Reflection (TIR)
Check both conditions for total internal reflection.Beyond the critical angle, light stays inside
Read the assumptions and limits
Total internal reflection requires light to go from higher to lower refractive index and incidence strictly greater than the critical angle. There is no propagating transmitted ray; an evanescent field can extend a short distance across the interface. The 90° prism and fibre paths are separate examples with different indices.
Original teaching diagrams · NCERT Physics Part II, chapter 9, PDF pages 9–12 · learner-controlled traces
- Critical angle
- The incidence angle in the higher-index medium for which the refracted ray grazes the interface: sin i_c = n₂/n₁, with n₁ > n₂.
Light must approach a lower-index medium from a higher-index one. Below the critical angle, reflection and refraction both occur; at the critical angle the transmitted direction is along the boundary. Above it, there is no propagating transmitted ray.
A right-angle glass prism can turn a ray by 90° through one total internal reflection. In an optical fibre, a higher-index core is surrounded by lower-index cladding, allowing suitable rays to be guided.
Go deeper: Useful reflection has physical limits
The critical condition is measured from the normal at each interface. Entry into a fibre does not guarantee guidance: the internal ray must satisfy the boundary condition.
An evanescent field can exist just outside a totally reflecting interface. Real fibres have absorption, scattering and bend losses; “total” describes the ideal interface reflection, not a loss-free cable of any length or bend.
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Question
A core has n₁ = 1.50 and cladding n₂ = 1.45. Is a ray incident internally at 78° guided by total internal reflection?
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Refraction at Spherical Surfaces and Lenses
Use the material contrast as well as the shape.Curvature and refractive index work together
Read the assumptions and limits
Thin lenses and small angles near the axis are assumed for image equations. Drawings are teaching schematics: instruments can use different horizontal and vertical scales. Solid paths show light travelling; dashed paths locate virtual images. Real devices have finite apertures and aberrations. D = 25 cm is a conventional reference near point, not a universal measurement for every eye.
Original teaching diagrams · NCERT Physics Part II, chapter 9, PDF pages 12–17 · learner-controlled traces
- Thin-lens equation
- 1/v − 1/u = 1/f. For a lens, m = v/u. A positive focal length means convergence in the stated surrounding medium.
For paraxial refraction at a spherical surface, n₂/v − n₁/u = (n₂−n₁)/R. The incident direction is positive; the sign of R follows the side of the centre of curvature.
For a thin lens in the same medium on both sides, 1/f = (n_lens/n_medium−1)(1/R₁−1/R₂). In air a biconvex glass lens normally converges. In a medium of higher index than the glass, the same shape can diverge.
Go deeper: How the lens equation is obtained
Apply the spherical-surface relation at each face. For a thin lens, the separation of the two vertices is negligible compared with the object and image distances. Combining the two relations gives the lens-maker equation and the thin-lens imaging equation.
Rays parallel to the axis pass through the far focus of a converging lens. A ray through the optical centre is approximately undeviated in the thin-lens model. A diverging lens requires backward extensions to locate its virtual image for a real incident object.
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Question
A glass biconvex lens has n = 1.50, R₁ = +20 cm and R₂ = −20 cm in air. Find its focal length. What if it is placed in an index-matched liquid?
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Combination of Thin Lenses
First change the ray slope, then carry that ray to the next lens.Two lenses share one ray path
Read the assumptions and limits
Thin lenses and small angles near the axis are assumed for image equations. Drawings are teaching schematics: instruments can use different horizontal and vertical scales. Solid paths show light travelling; dashed paths locate virtual images. Real devices have finite apertures and aberrations. D = 25 cm is a conventional reference near point, not a universal measurement for every eye.
Original teaching diagrams · NCERT Physics Part II, chapter 9, PDF pages 17–19 · learner-controlled traces
- Optical power
- P = 1/f in metres, measured in dioptres (D). Thin lenses in contact have P_total = P₁ + P₂.
For thin lenses in contact, adding their powers gives the effective focal length. Positive and negative powers can reinforce or cancel. Zero net power means no convergence of a parallel bundle in the ideal contact model.
A separated pair cannot generally use the contact formula. In air, the paraxial equivalent power is P₁ + P₂ − dP₁P₂, where d is separation in metres. The ray height at the second lens depends on its travel from the first.
Go deeper: Objects for the second lens
To locate an actual final image, apply the first lens equation, then measure the intermediate image position relative to the second lens. Rays converging towards a point beyond the second lens give it a virtual object with u₂ > 0.
For separated systems the principal planes also shift. An equivalent focal length alone does not locate every image measured from a lens vertex. The separated-power formula is a supporting paraxial extension, while contact addition and successive imaging are the core chapter methods.
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Question
A +4 D lens and a −1.5 D lens are in contact. Find the combined power and focal length.
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Refraction Through a Prism
Trace both interfaces before adding their directional changes.Two refractions make one deviation
Read the assumptions and limits
The prism angle is 60°. Angles are measured from the normals. Ordinary normal dispersion is illustrated; the five colours use illustrative indices 1.46–1.54, exaggerating colour separation; this is not a measured spectrum of a particular glass. At minimum deviation the internal ray is parallel to the base here because this prism is isosceles. That statement is not true for every triangular prism.
Original teaching diagrams · NCERT Physics Part II, chapter 9, PDF pages 19–20 · learner-controlled traces
- Angle of deviation
- δ = i + e − A for a transmitted ray through a prism; the internal refraction angles satisfy r₁ + r₂ = A.
The entry and exit angles are measured from different face normals. Use Snell’s law at the first face to find r₁, geometry to get r₂ = A−r₁, and Snell’s law at the second face to find e.
At minimum deviation, the path is symmetric: i = e and r₁ = r₂ = A/2. Therefore the relative index is sin((A+δ_m)/2)/sin(A/2). For a thin prism with small angles, δ ≈ (n−1)A.
Go deeper: Colour and geometry
In normal dispersion, refractive index is greater for violet than red, so violet is deviated further towards the base. Refraction changes wavelength inside the prism; the frequency remains unchanged throughout.
In an isosceles prism the minimum-deviation internal ray is parallel to the base. Symmetry does not imply this for every triangular prism. Some incident angles make the second-face incidence exceed the critical angle, so an emerging ray cannot be assumed.
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Question
A 60° prism has minimum deviation 40°. Calculate its refractive index relative to air.
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The Microscope
Let the objective make an image that becomes the eyepiece’s object.First a real image, then angular enlargement
Read the assumptions and limits
Thin lenses and small angles near the axis are assumed for image equations. Drawings are teaching schematics: instruments can use different horizontal and vertical scales. Solid paths show light travelling; dashed paths locate virtual images. Real devices have finite apertures and aberrations. D = 25 cm is a conventional reference near point, not a universal measurement for every eye.
Original teaching diagrams · NCERT Physics Part II, chapter 9, PDF pages 20–24 · learner-controlled traces
- Angular magnification
- The angle subtended at the eye through the instrument divided by the reference angle for unaided viewing at the near point D.
A simple magnifier places the object within a converging lens’s focal length. For an eye close to the lens, angular magnification is D/f with the final image at infinity, or 1+D/f with the final image at the near point. D = 25 cm is the conventional reference.
A compound microscope’s short-focal-length objective forms a real, inverted enlarged intermediate image. The eyepiece acts as a magnifier. The total angular magnification magnitude is the objective’s transverse magnification magnitude times the eyepiece’s angular magnification.
Go deeper: Relaxed-eye and near-point arrangements
For an objective with fₒ = 5 cm and uₒ = −6 cm, vₒ = +30 cm and mₒ = −5. An eyepiece with fₑ = 10 cm gives magnitude 2.5 for a relaxed eye or 3.5 at the 25 cm reference near point. The complete instrument magnitudes are 12.5 and 17.5 respectively.
The common relaxed-eye approximation |M| ≈ (L/fₒ)(D/fₑ) uses optical tube length L between the objective’s far focus and the eyepiece’s near focus, not simply any measured barrel length. More magnification does not overcome diffraction or poor illumination.
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Question
A microscope objective gives transverse magnification −8. Its eyepiece has f = 5 cm and is adjusted for a relaxed eye. With D = 25 cm, find the total angular magnification and orientation.
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The Telescope
Compare angle enlargement, light collection and a folded reflecting path.A large objective collects light; an eyepiece enlarges the angle
Read the assumptions and limits
Thin lenses and small angles near the axis are assumed for image equations. Drawings are teaching schematics: instruments can use different horizontal and vertical scales. Solid paths show light travelling; dashed paths locate virtual images. Real devices have finite apertures and aberrations. D = 25 cm is a conventional reference near point, not a universal measurement for every eye.
Original teaching diagrams · NCERT Physics Part II, chapter 9, PDF pages 24–26 · learner-controlled traces
- Normal adjustment
- A refracting telescope setting with both object and final image at infinity. Objective and eyepiece separation is fₒ + fₑ and angular magnification is −fₒ/fₑ.
The objective forms a real image of a distant scene near its focal plane. The eyepiece’s near focal plane coincides with this image, making each outgoing ray bundle parallel for relaxed viewing. The negative angular magnification gives an inverted view.
A larger objective aperture gathers more light and can improve diffraction-limited resolution. Reflecting telescopes use mirrors: a Cassegrain arrangement sends light from a concave primary to a convex secondary and back through a central opening in the primary.
Go deeper: What the telescope improves
Light collection for a circular clear aperture scales with diameter squared, with other conditions fixed. Resolution depends on aperture, wavelength and optical quality; increasing the eyepiece’s magnification alone cannot reveal unlimited new detail.
Mirrors avoid the wavelength-dependent refraction that gives a lens chromatic aberration. Reflectors can still have other aberrations, alignment errors and secondary obstruction. The fold reduces instrument length; it does not shorten the optical path travelled by light.
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Question
A normally adjusted telescope has fₒ = 80 cm and fₑ = 8 cm. Find magnification and separation. What happens to collected light if its unobstructed objective diameter doubles?
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