Ray Optics and Optical Instruments Class 12: Notes, Derivations and Numericals

What Ray Optics and Optical Instruments is about

Weightage: This chapter is part of Units V and VI, Electromagnetic Waves; Optics, which carry 18 of the 70 theory marks together with Electromagnetic Waves and Wave Optics.Source: CBSE Physics syllabus 2026-27, Units V and VI Electromagnetic Waves; Optics

In real board papers: asked in 8 of the 8 board papers we hold (2023 to 2026), on average 10.9 marks per paper. See the questions

Key concepts

Spherical mirrors and the sign convention

NCERT section 9.2.1 and 9.2.2

Cartesian sign convention (used throughout NCERT): all distances are measured from the pole (mirror) or optical centre (lens). Distances in the direction of the incident light are positive, against it negative. Heights above the principal axis are positive, below negative. With light coming from the left, a real object always has a negative u.

So a concave mirror has negative f and a convex mirror positive f; a convex lens has positive f and a concave lens negative f.

Focal length: for a spherical mirror of small aperture, paraxial rays parallel to the axis meet (or appear to diverge from) the focus F midway between the pole and the centre of curvature: $f=\frac{R}{2}$.

Worked example

A concave mirror has a radius of curvature of 30 cm. Write its focal length with sign. Where must an object be placed to get an image of the same size? [1 mark]

Try it first, then show the solutionSolution

1. $f=\frac{R}{2}=-\frac{30}{2}=-15cm$ (concave: negative)
2. Same-size image when the object is at C: u = −30 cm (image also at C, real and inverted).

Answer: f = −15 cm; place the object at the centre of curvature, 30 cm in front.

The mirror equation and magnification

NCERT section 9.2.3

Derivation (concave mirror, real image). An object AB at distance u forms an image A'B' at v. A ray from B parallel to the axis reflects through F; a ray from B through F reflects parallel. From similar triangles A'B'F and MPF (M on the mirror, P the pole, using PM ≈ AB for small aperture): $\frac{B^{\prime }{A}^{\prime }}{BA}=\frac{F{A}^{\prime }}{FP}$. From triangles A'B'P and ABP (the ray to the pole reflects at equal angles): $\frac{B^{\prime }{A}^{\prime }}{BA}=\frac{P{A}^{\prime }}{PA}$. So $\frac{F{A}^{\prime }}{FP}=\frac{P{A}^{\prime }}{PA}$, i.e. $\frac{P{A}^{\prime }-PF}{PF}=\frac{P{A}^{\prime }}{PA}$. Substituting with signs, $P{A}^{\prime }=-v$, $PF=-f$, $PA=-u$: $\frac{-v+f}{-f}=\frac{v}{u}$, which rearranges to

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$

Magnification: $m=\frac{h^{\prime }}{h}=-\frac{v}{u}$. Negative m means an inverted (real) image; positive m an erect (virtual) image; |m| > 1 means magnified. A convex mirror always gives a virtual, erect, diminished image.

Worked example

An object is 15 cm in front of a concave mirror of focal length 10 cm. Find the position, nature and magnification of the image. [2 marks]

Try it first, then show the solutionSolution

1. u = −15 cm, f = −10 cm: $\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=-\frac{1}{10}+\frac{1}{15}=\frac{-3+2}{30}=-\frac{1}{30}$
2. v = −30 cm: the image is 30 cm in front of the mirror (real)
3. $m=-\frac{v}{u}=-\frac{-30}{-15}=-2$: inverted and twice the size

Answer: Real, inverted image 30 cm in front of the mirror, magnification −2

Refraction and apparent depth

NCERT section 9.3

Snell's law: $n_1\sin i=n_2\sin r$, or $\frac{\sin i}{\sin r}=n_{21}=\frac{n_2}{n_1}$, the refractive index of medium 2 relative to medium 1. Also $n_{21}=\frac{v_1}{v_2}$ and $n_{12}=\frac{1}{n_{21}}$. Light bends towards the normal on entering a denser medium. Its frequency is unchanged; its speed and wavelength decrease.

A parallel-sided slab shifts a ray sideways (lateral shift) without changing its direction.

Apparent depth: an object at real depth h under a liquid of refractive index n, viewed from nearly above, appears raised to $h^{\prime }=\frac{h}{n}$. The apparent rise is $h\left(1-\frac{1}{n}\right)$.

Worked example

A coin lies at the bottom of a tank filled with water ($n=\frac{4}{3}$) to a depth of 12 cm. Viewed from above, how deep does the coin appear and by how much is it raised? [1 mark]

Try it first, then show the solutionSolution

1. $h^{\prime }=\frac{h}{n}=\frac{12}{4/3}=9cm$
2. Raised by $12-9=3cm$

Answer: It appears 9 cm deep, raised by 3 cm.

Total internal reflection and optical fibres

NCERT section 9.4

When light goes from a denser to a rarer medium, the refracted ray bends away from the normal. At a particular angle of incidence, the critical angle $i_c$, the refracted ray grazes the surface ($r=90^{\circ }$): $\sin i_c=\frac{n_2}{n_1}$, or $\sin i_c=\frac{1}{n}$ for a medium of index n in air. For larger angles no light is refracted and all of it is reflected: total internal reflection (TIR).

Two conditions: light must travel from a denser to a rarer medium, and the angle of incidence must exceed the critical angle.

Uses:

• Totally reflecting prisms (right-angled isosceles prisms, angles 45°) turn rays through 90° or 180°, or invert images, without the loss of a silvered mirror, provided the critical angle of the glass is below 45°.
• Optical fibres: a thin glass or plastic core is coated with cladding of lower refractive index. Light entering at a suitable angle strikes the core-cladding boundary at more than the critical angle and is totally reflected again and again along the fibre, even round gentle bends, with very little loss. Used to carry telephone and internet signals and in medical endoscopes.

Worked example

An optical fibre has a core of refractive index 1.60 and a cladding of refractive index 1.50. Find the critical angle at the core-cladding boundary. What happens to a ray that strikes the boundary at $65^{\circ }$? [2 marks]

Try it first, then show the solutionSolution

1. $\sin i_c=\frac{n_{cladding}}{n_{core}}=\frac{1.50}{1.60}=0.9375$
2. $i_c\approx 69.6^{\circ }$
3. $65^{\circ }<i_c$, so the ray is partly refracted into the cladding and is not totally reflected: light leaks out.

Answer: About 69.6°; at 65° the ray is not totally reflected and leaks into the cladding.

Refraction at a spherical surface

NCERT section 9.5.1

Derivation. A point object O in medium $n_1$ is on the axis of a convex spherical surface (centre C, radius R) bounding medium $n_2$; the image is I. A ray from O meets the surface at N, making angle i with the normal NC, and refracts at angle r. For small angles (paraxial rays), with NM the perpendicular height: $\tan \angle NOM\approx \frac{MN}{OM}$, $\tan \angle NCM\approx \frac{MN}{MC}$, $\tan \angle NIM\approx \frac{MN}{MI}$.

From triangle NOC: $i=\angle NOM+\angle NCM$. From triangle NCI: $r=\angle NCM-\angle NIM$. Snell's law for small angles: $n_{1}i=n_{2}r$. Substituting:

$\frac{n_1}{OM}+\frac{n_2}{MI}=\frac{n_2-n_1}{MC}$

With signs OM = −u, MI = +v, MC = +R:

$\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R}$

This single result, applied twice, gives the lens formulas.

Worked example

A point object in air is 100 cm in front of the convex surface of a long glass rod (n = 1.5) whose end has a radius of curvature of 20 cm. Where is the image formed? [2 marks]

Try it first, then show the solutionSolution

1. $n_1=1$, $n_2=1.5$, u = −100 cm, R = +20 cm
2. $\frac{1.5}{v}=\frac{1.5-1}{20}+\frac{1}{-100}=0.025-0.01=0.015$
3. $v=\frac{1.5}{0.015}=100cm$: a real image inside the glass, 100 cm from the surface

Answer: Real image 100 cm inside the rod

Lens maker's formula

NCERT section 9.5.2

Derivation. A thin lens of index $n_2$ in a medium of index $n_1$ has two surfaces of radii $R_1$ and $R_2$. The first surface forms an image $I_1$ of the object O: $\frac{n_2}{v_1}-\frac{n_1}{u}=\frac{n_2-n_1}{R_1}$. This image acts as the object for the second surface, which forms the final image I: $\frac{n_1}{v}-\frac{n_2}{v_1}=\frac{n_1-n_2}{R_2}$ (thin lens, so distances are measured from the same point). Adding:

$\frac{n_1}{v}-\frac{n_1}{u}=(n_2-n_1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

For an object at infinity, v = f, so with $n_{21}=\frac{n_2}{n_1}$:

$\frac{1}{f}=(n_{21}-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

This is the lens maker's formula. Signs: for a double convex lens, $R_1>0$ and $R_2<0$. In a denser surrounding medium $n_{21}$ falls, so f increases; if the surrounding medium is denser than the lens ($n_{21}<1$), a convex lens diverges.

Worked example

An equiconvex lens of glass (n = 1.5) has radii of curvature of 20 cm. Find its focal length (a) in air, (b) in water ($n_w=\frac{4}{3}$). [2 marks]

Try it first, then show the solutionSolution

1. $R_1=+20cm$, $R_2=-20cm$, so $\frac{1}{R_1}-\frac{1}{R_2}=\frac{2}{20}=\frac{1}{10}$
2. (a) $\frac{1}{f}=(1.5-1)\times \frac{1}{10}=\frac{1}{20}$, so f = 20 cm
3. (b) $n_{21}=\frac{1.5}{4/3}=1.125$: $\frac{1}{f^{\prime }}=0.125\times \frac{1}{10}=\frac{1}{80}$, so f' = 80 cm

Answer: (a) 20 cm (b) 80 cm, four times longer

Thin lens formula and magnification

NCERT section 9.5.2

Putting $\frac{1}{f}$ from the lens maker's formula into the two-surface result gives the thin lens formula:

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

with the Cartesian sign convention (u negative for a real object). The magnification is $m=\frac{h^{\prime }}{h}=\frac{v}{u}$: negative for a real inverted image, positive for a virtual erect one.

A convex lens gives a real image when the object is beyond F, and a virtual magnified image when the object is between F and the lens. A concave lens always gives a virtual, erect, diminished image. Covering half of a lens still gives the whole image, only dimmer, because every part of the lens forms the complete image.

Worked example

An object 2 cm tall is placed 30 cm from a convex lens of focal length 20 cm. Find the position, size and nature of the image. [2 marks]

Try it first, then show the solutionSolution

1. u = −30 cm, f = +20 cm: $\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{20}-\frac{1}{30}=\frac{1}{60}$, so v = +60 cm
2. $m=\frac{v}{u}=\frac{60}{-30}=-2$
3. $h^{\prime }=mh=-4cm$: real, inverted, 4 cm tall, 60 cm behind the lens

Answer: Real, inverted, 4 cm tall, 60 cm on the other side of the lens

Power of a lens and lenses in contact

NCERT section 9.5.3 and 9.5.4

The power of a lens measures how strongly it converges or diverges light: $P=\frac{1}{f}$ with f in metres. Unit: dioptre (D) = m⁻¹. Converging lenses have positive power, diverging lenses negative.

Thin lenses in contact. The first lens forms an image at $v_1$: $\frac{1}{v_1}-\frac{1}{u}=\frac{1}{f_1}$. This acts as the object for the second: $\frac{1}{v}-\frac{1}{v_1}=\frac{1}{f_2}$. Adding: $\frac{1}{v}-\frac{1}{u}=\frac{1}{f_1}+\frac{1}{f_2}$. So the combination behaves like one lens with

$\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}+\dots ,P=P_1+P_2+\dots$

The total magnification of a combination is the product $m=m_1{m}_2{m}_3\dots$. Opticians prescribe lenses by power because powers simply add.

Worked example

A convex lens of focal length 20 cm is placed in contact with a concave lens of focal length 40 cm. Find the power and focal length of the combination. Is it converging or diverging? [2 marks]

Try it first, then show the solutionSolution

1. $P_1=\frac{1}{0.20}=+5D$; $P_2=\frac{1}{-0.40}=-2.5D$
2. $P=5-2.5=+2.5D$
3. $f=\frac{1}{2.5}=0.4m=40cm$, converging

Answer: +2.5 D; f = +40 cm; converging

Refraction through a prism

NCERT section 9.6

A ray enters face AB at angle i, refracts at $r_1$, meets face AC at $r_2$ and leaves at angle e. Geometry of the quadrilateral formed by the normals gives

$r_1+r_2=A$

The total deviation is $\delta =(i-r_1)+(e-r_2)=i+e-A$.

As i increases, δ first decreases, reaches a minimum $D_m$ and then increases. At minimum deviation the ray passes symmetrically through the prism: $i=e$ and $r_1=r_2=\frac{A}{2}$; also $i=\frac{A+D_m}{2}$. Snell's law at the first face then gives

$n_{21}=\frac{\sin \left(\frac{A+D_m}{2}\right)}{\sin \left(\frac{A}{2}\right)}$

For a thin prism (small A), $D_m=(n_{21}-1)A$.

Worked example

The angle of minimum deviation of a prism of angle $60^{\circ }$ is $30^{\circ }$. Find its refractive index and the angle of incidence at minimum deviation. [2 marks]

Try it first, then show the solutionSolution

1. $n=\frac{\sin \left(\frac{60^{\circ }+30^{\circ }}{2}\right)}{\sin 30^{\circ }}=\frac{\sin 45^{\circ }}{0.5}=\frac{0.707}{0.5}$
2. $n=\sqrt{2}\approx 1.414$
3. $i=\frac{A+D_m}{2}=45^{\circ }$

Answer: $n=\sqrt{2}\approx 1.41$; i = 45°

Simple and compound microscopes

NCERT section 9.7.1

Simple microscope (a single converging lens of focal length f). With the image at the near point D (about 25 cm), the magnification is $m=1+\frac{D}{f}$; with the image at infinity, $m=\frac{D}{f}$. A shorter focal length gives more magnification.

Compound microscope: an objective of short focal length $f_o$ forms a real, inverted, magnified image of the object near the focus of the eyepiece; the eyepiece (focal length $f_e$) acts as a simple microscope and magnifies this image further. The total magnification is $m=m_o{m}_e$.

• Objective: $m_o=\frac{L}{f_o}$, where L (the tube length) is the distance between the second focal point of the objective and the first focal point of the eyepiece.
• Final image at infinity: $m=\frac{L}{f_o}\times \frac{D}{f_e}$.
• Final image at the near point: $m=\frac{L}{f_o}\left(1+\frac{D}{f_e}\right)$, or more exactly $m_o\times \left(1+\frac{D}{f_e}\right)$ with $m_o=\frac{v_o}{|u_o|}$.

Both lenses should have short focal lengths, the objective the shorter.

Worked example

A compound microscope has an objective of focal length 1 cm and an eyepiece of focal length 5 cm, with tube length 15 cm. Find its magnifying power when the final image is at infinity (D = 25 cm). [2 marks]

Try it first, then show the solutionSolution

1. $m_o=\frac{L}{f_o}=\frac{15}{1}=15$
2. $m_e=\frac{D}{f_e}=\frac{25}{5}=5$
3. $m=15\times 5=75$

Answer: 75

Refracting and reflecting telescopes

NCERT section 9.7.2

Refracting (astronomical) telescope: an objective of long focal length $f_o$ and large aperture forms a real, inverted image of a distant object at its focus; an eyepiece of short focal length $f_e$ magnifies it. In normal adjustment (final image at infinity) the magnifying power is the ratio of the angles subtended at the eye and at the objective:

$m=\frac{\beta }{\alpha }=\frac{f_o}{f_e}$, and the length of the tube is $f_o+f_e$

With the final image at the near point, $m=\frac{f_o}{f_e}\left(1+\frac{f_e}{D}\right)$. A large objective aperture gathers more light and resolves finer detail.

Reflecting telescope (for example the Cassegrain type, with a concave primary mirror and a convex secondary mirror that sends the light through a hole in the primary to the eyepiece). Advantages over large refractors: no chromatic aberration (a mirror reflects all colours alike); a paraboloidal mirror removes spherical aberration; a mirror is much lighter than a lens of the same aperture and can be supported over its whole back, while a big lens can be held only at its rim.

Worked example

An astronomical telescope has an objective of focal length 150 cm and an eyepiece of focal length 5 cm. Find its magnifying power and length in normal adjustment. [1 mark]

Try it first, then show the solutionSolution

1. $m=\frac{f_o}{f_e}=\frac{150}{5}=30$
2. Length $=f_o+f_e=155cm$

Answer: Magnifying power 30; length 155 cm

Physics in real life: Ray Optics and Optical Instruments

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Total Internal Reflection
• Refraction at Spherical Surfaces and Lenses
• Microscopes and Telescopes

Ray Optics and Optical Instruments formula sheet

Focal length of a mirror

$f=\frac{R}{2}$

Mirror equation

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$

Cartesian signs; concave f < 0.

Magnification (mirror)

$m=\frac{h^{\prime }}{h}=-\frac{v}{u}$

Snell's law

$n_1\sin i=n_2\sin r,n_{21}=\frac{v_1}{v_2}$

Apparent depth

$h^{\prime }=\frac{h}{n}$

Critical angle

$\sin i_c=\frac{n_2}{n_1}=\frac{1}{n}\text{(into air)}$

Refraction at a spherical surface

$\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R}$

Lens maker's formula

$\frac{1}{f}=(n_{21}-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

Thin lens formula

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

Magnification (lens)

$m=\frac{h^{\prime }}{h}=\frac{v}{u}$

Power and lenses in contact

$P=\frac{1}{f(m)},P=P_1+P_2,m=m_1{m}_2$

Prism

$r_1+r_2=A,\delta =i+e-A$

Refractive index of a prism

$n_{21}=\frac{\sin \left(\frac{A+D_m}{2}\right)}{\sin \frac{A}{2}}$

Thin prism: $D_m=(n-1)A$.

Simple microscope

$m=1+\frac{D}{f}\text{(near point)},m=\frac{D}{f}\text{(infinity)}$

Compound microscope

$m=\frac{L}{f_o}\cdot \frac{D}{f_e}\text{(infinity)},m=\frac{L}{f_o}\left(1+\frac{D}{f_e}\right)\text{(near point)}$

Astronomical telescope

$m=\frac{f_o}{f_e},L=f_o+f_e\text{(normal adjustment)}$

Near point: $m=\frac{f_o}{f_e}\left(1+\frac{f_e}{D}\right)$.

Ray Optics and Optical Instruments: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the board asks Ray Optics and Optical Instruments

The question patterns to expect, and the method that works for each.

Tricks to remember Ray Optics and Optical Instruments

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Mirror adds, lens subtracts

  Mirror: 1/v PLUS 1/u = 1/f. Lens: 1/v MINUS 1/u = 1/f. Magnification: mirror −v/u, lens +v/u.

  Use it when: Avoids mixing the two formulas.

• Signs follow the light

  Distances in the direction the light travels are positive. A real object is always on the incoming side, so u is negative.

  Use it when: Every numerical.

• Objective small for microscope, big for telescope

  Microscope: tiny object, tiny $f_o$. Telescope: far object, huge $f_o$. Eyepiece is short in both.

  Use it when: Instrument design questions.

• TIR: Dense to Rare, and Beyond Critical

  DRBC: Denser to Rarer, Beyond the Critical angle.

  Use it when: Conditions for TIR.

• Prism at minimum deviation is symmetric

  At $D_m$: i = e, $r_1=r_2=A/2$. Then $n=\sin \frac{A+D_m}{2}/\sin \frac{A}{2}$.

  Use it when: Prism derivation and numericals.

• Water stretches a lens

  In water a glass lens's focal length grows (roughly four times for n = 1.5): the lens 'weakens'.

  Use it when: Lens-in-liquid MCQs.

Ray Optics and Optical Instruments: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section D style)

Read the situation, then answer every part. In the board paper this is a 4-mark question split into 1 + 1 + 1 + 1 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Ray Optics and Optical Instruments

The slips examiners see most often, and what to write instead.

• ✗Mistake: Dropping the signs of u, f or R

  Using magnitudes and guessing the answer's sign.

  Wrong working

  Concave mirror f = 10, u = 15: 1/v = 1/10 − 1/15

  Right working

  f = −10, u = −15: $\frac{1}{v}=-\frac{1}{10}+\frac{1}{15}$, v = −30 cm

  ✓Do this: Write 'u = −15 cm, f = −10 cm' before substituting.

  Usually costs: Usually all numerical marks.

• ✗Mistake: Using the mirror magnification for a lens

  Remembering only one formula.

  ✓Do this: Mirror: m = −v/u. Lens: m = +v/u.

  Usually costs: 1 mark.

• ✗Mistake: Forgetting $n_{21}$ is relative in the lens maker's formula

  Using the lens's absolute index in water.

  Wrong working

  Lens in water: $\frac{1}{f}=(1.5-1)(\dots )$

  Right working

  $\frac{1}{f}=\left(\frac{1.5}{1.33}-1\right)(\dots )$

  ✓Do this: Divide the lens index by the medium index.

  Usually costs: 1-2 marks.

• ✗Mistake: Ray diagram of a microscope without the intermediate image

  Drawing only the lenses.

  ✓Do this: Show the objective's real, inverted, magnified image A'B' just inside the eyepiece's focus, then the eyepiece's virtual final image.

  Usually costs: 1-2 marks of the diagram.

• ✗Mistake: Writing power in dioptres with f in cm

  Forgetting units.

  Wrong working

  f = 25 cm: P = 1/25 = 0.04 D

  Right working

  P = 1/0.25 m = 4 D

  ✓Do this: Convert f to metres before P = 1/f.

  Usually costs: 1 mark.

• ✗Mistake: Explaining mirages or rainbows as TIR applications

  Older syllabi included them.

  ✓Do this: These are not in the 2026-27 syllabus; for TIR applications use optical fibres and totally reflecting prisms.

  Usually costs: Marks if the expected examples are missing.

NCERT exercise guide: Ray Optics and Optical Instruments

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Ray Optics and Optical Instruments: previous year questions

• Asked in 8 of the 8 board papers we hold (2023 to 2026), on average 10.9 marks per paper; 32 questions in all, counting OR alternatives. In 4 of 4 sample papers.

Counted from the real papers on our previous year papers page, not estimated. Average marks are over every board paper we hold, with an OR pair counted once. Chapter tags are ours.

Solved board questions from Ray Optics and Optical Instruments

Real questions from recent papers. Try each one first, then open our step-by-step solution with marks per step.

Every Ray Optics and Optical Instruments question, by year

Practice questions on Ray Optics and Optical Instruments

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQBoard level

   The critical angle for a medium of refractive index $\sqrt{2}$ is:
   (a) 30°
   (b) 45°
   (c) 60°
   (d) 90°

   Hint

   $\sin i_c=1/n$.

   Answer

   (b) 45°

2. Q2 · 1 mark · MCQBoard level

   An object is placed at the centre of curvature of a concave mirror. The image is:
   (a) at F, diminished
   (b) at C, same size, inverted
   (c) behind the mirror, erect
   (d) at infinity

   Hint

   u = 2f gives v = 2f.

   Answer

   (b)

3. Q3 · 1 mark · Very short answerBoard level

   State the two conditions necessary for total internal reflection.

   Hint

   Direction of travel and angle.

   Answer

   Light must travel from a denser to a rarer medium, and the angle of incidence must be greater than the critical angle.

4. Q4 · 1 mark · Assertion-reasonHarder board level

   Assertion (A): The focal length of a convex glass lens increases when it is immersed in water.
   Reason (R): The refractive index of glass relative to water is less than that relative to air.
   (a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.

   Hint

   Lens maker's formula.

   Answer

   (a)

5. Q5 · 2 marks · Very short answerHarder board level

   An object is 5 cm in front of a concave mirror of focal length 10 cm. Find the position, magnification and nature of the image.

   Hint

   Object inside F.

   Answer

   $\frac{1}{v}=-\frac{1}{10}+\frac{1}{5}=\frac{1}{10}$, v = +10 cm (behind the mirror); $m=-\frac{10}{-5}=+2$: virtual, erect, twice the size.

6. Q6 · 2 marks · Very short answerHarder board level

   A small lamp at the bottom of a tank of liquid of refractive index $\frac{5}{3}$ lights up a circle of radius 45 cm on the surface; outside this circle no light escapes. (a) Find the critical angle. (b) Find the depth of the liquid. (c) Some liquid is drained so that the level falls by 20 cm. Find the new radius of the circle.

   Hint

   Light escapes only where it meets the surface at less than the critical angle: $r=h\tan i_c$.

   Answer

   (a) $\sin i_c=\frac{3}{5}=0.6$, so $i_c\approx 36.9^{\circ }$ and $\tan i_c=\frac{0.6}{0.8}=0.75$. (b) $h=\frac{r}{\tan i_c}=\frac{45}{0.75}=60cm$. (c) New depth 40 cm, so $r=40\times 0.75=30cm$.

7. Q7 · 3 marks · Short answerHarder board level

   Thin lenses of power +6 D and −2 D are placed in contact. (a) An object 2 cm tall is 75 cm from the combination. Find the position, size and nature of the image. (b) The −2 D lens is replaced by another lens so that the image of the same object forms 50 cm behind the combination. What is the power of the new lens?

   Hint

   Powers add; then use $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ with f in cm.

   Answer

   (a) P = +4 D, f = +25 cm. u = −75 cm: $\frac{1}{v}=\frac{1}{25}-\frac{1}{75}=\frac{2}{75}$, v = +37.5 cm. $m=\frac{37.5}{-75}=-0.5$: real, inverted, 1 cm tall. (b) Needed $\frac{1}{f}=\frac{1}{50}+\frac{1}{75}=\frac{1}{30}$, so P = $\frac{100}{30}\approx 3.33$ D; new lens power $=3.33-6\approx -2.67$ D.

8. Q8 · 3 marks · Short answerHarder board level

   An astronomical telescope in normal adjustment is 105 cm long and has a magnifying power of 20. (a) Find the focal lengths of its objective and eyepiece. (b) Find its magnifying power when the final image is at 25 cm. (c) The Moon subtends about 0.5° at the objective. What angle does the final image subtend at the eye in normal adjustment?

   Hint

   $f_o+f_e=105$ and $f_o/f_e=20$.

   Answer

   (a) $f_o=20f_e$ and $21f_e=105$, so $f_e=5cm$, $f_o=100cm$. (b) $m=20\left(1+\frac{5}{25}\right)=24$. (c) $\beta =m\alpha =20\times 0.5^{\circ }=10^{\circ }$.

9. Q9 · 5 marks · Long answerHarder board level

   (a) Draw a labelled ray diagram of a compound microscope with the final image at the near point and derive its magnifying power. (b) A microscope has $f_o=1.5cm$ and $f_e=5cm$, and an object is 1.8 cm from the objective. Find the magnifying power and the lens separation for a final image at 25 cm.

   Hint

   $m=m_o{m}_e$.

   Answer

   (a) $m=\frac{v_o}{|u_o|}\left(1+\frac{D}{f_e}\right)$. (b) $\frac{1}{v_o}=\frac{1}{1.5}-\frac{1}{1.8}=\frac{1}{9}$, so $v_o=9cm$ and $m_o=\frac{9}{1.8}=5$; $m_e=1+\frac{25}{5}=6$; m = 30. Eyepiece: $\frac{1}{u_e}=-\frac{1}{25}-\frac{1}{5}=-\frac{6}{25}$, $u_e\approx -4.17cm$; separation $\approx 9+4.17=13.2cm$.

10. Q10 · 3 marks · Short answerHOTS / competency

    A convex lens of focal length 20 cm is moved between an object and a screen that are 90 cm apart. (a) Find the two positions of the lens that give a sharp image on the screen. (b) How far apart are they? (c) Find the ratio of the two image sizes.

    Hint

    With |u| + v = 90 and $\frac{1}{v}+\frac{1}{|u|}=\frac{1}{20}$, |u| and v are the roots of a quadratic.

    Answer

    (a) $\frac{|u|+v}{|u|v}=\frac{1}{20}$ gives $|u|v=1800$ with $|u|+v=90$, so |u| and v are the roots of $x^2-90x+1800=0$: 30 and 60. The lens can be 30 cm or 60 cm from the object. (b) 30 cm apart. (c) Magnifications 60/30 = 2 and 30/60 = 1/2, so the image sizes are in the ratio 4 : 1.

11. Q11 · 2 marks · Very short answerHOTS / competency

    A thin prism of angle 5° and refractive index 1.6 is combined with another thin prism of refractive index 1.5 so that there is no net deviation. Find the angle of the second prism.

    Hint

    Thin prism: δ = (n − 1)A; place them inverted.

    Answer

    $(1.6-1)\times 5=(1.5-1)A^{\prime }$, so $A^{\prime }=\frac{3}{0.5}=6^{\circ }$ (placed with its apex the other way).

12. Q12 · 1 mark · Very short answerHarder board level

    Half of a convex lens is covered with black paper. How does the image of an object change?

    Hint

    Each part of the lens forms a full image.

    Answer

    The full image is still formed at the same place but it is less bright (about half the light).

13. Q13 · 2 marks · Very short answerHarder board level

    A planoconvex lens has focal length 30 cm and is made of glass of n = 1.5. Find the radius of curvature of its curved surface. What would its focal length be if made of glass of n = 1.6?

    Hint

    Plane surface: $1/R=0$.

    Answer

    $\frac{1}{30}=0.5\times \frac{1}{R}$, so R = 15 cm. With n = 1.6: $\frac{1}{f}=0.6\times \frac{1}{15}$, f = 25 cm.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.Write the mirror equation and the lens formula.Show answer

   Mirror: $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$. Lens: $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$.

2. 2.Two conditions for total internal reflection?Show answer

   Light goes from a denser to a rarer medium, and the angle of incidence exceeds the critical angle.

3. 3.Critical angle for glass of n = 1.5?Show answer

   $\sin i_c=1/1.5$, $i_c\approx 41.8^{\circ }$.

4. 4.Write the lens maker's formula.Show answer

   $\frac{1}{f}=(n_{21}-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$.

5. 5.Power of a lens of focal length −50 cm?Show answer

   −2 D.

6. 6.Formula for the refractive index of a prism?Show answer

   $n=\frac{\sin \frac{A+D_m}{2}}{\sin \frac{A}{2}}$.

7. 7.Magnifying power of an astronomical telescope in normal adjustment?Show answer

   $m=f_o/f_e$; length $f_o+f_e$.

8. 8.Give two advantages of a reflecting telescope.Show answer

   No chromatic aberration; spherical aberration removed with a paraboloidal mirror; lighter and easier to support (any two).

9. 9.Magnifying power of a compound microscope with image at infinity?Show answer

   $m=\frac{L}{f_o}\cdot \frac{D}{f_e}$.