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

What Ray Optics and Optical Instruments is about

Weightage: Ray Optics and Optical Instruments is part of the unit Optics, which carries 18 of the 70 marks in the ISC Physics Paper 1 (Theory), shared with 1 other chapter. Questions on it can appear in any section of the paper.Source: CISCE ISC Physics syllabus, examination year 2027, distribution of marks

Key concepts

Spherical mirrors and the mirror formula

For a spherical mirror the focal length is half the radius of curvature, $f=R/2$ (paraxial rays). Using similar triangles in the ray diagram of a concave mirror forming a real image (object AB, image A'B', pole P, centre C): $\frac{A^{\prime }{B}^{\prime }}{AB}=\frac{B^{\prime }C}{BC}=\frac{B^{\prime }F}{FP}$, which leads to $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ (with signs). Linear magnification $m=\frac{h^{\prime }}{h}=-\frac{v}{u}$: negative for real inverted images, positive for virtual erect ones. With the New Cartesian convention, a concave mirror has negative f and a convex mirror positive f. The focal length of a mirror does not depend on the surrounding medium.

Worked example

An object is placed 25 cm in front of a concave mirror of focal length 15 cm. Find the position, nature and magnification of the image.

Try it first, then show the solutionSolution

1. u = −25 cm, f = −15 cm. $\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=-\frac{1}{15}+\frac{1}{25}=-\frac{2}{75}$
2. v = −37.5 cm (in front of the mirror: real)
3. $m=-\frac{v}{u}=-\frac{-37.5}{-25}=-1.5$ (inverted, magnified)

Answer: 37.5 cm in front of the mirror; real, inverted, 1.5 times magnified

Refraction at plane surfaces and total internal reflection

Snell's law: $n_1\sin i=n_2\sin r$, i.e. ${}_1{n}_2=\frac{\sin i}{\sin r}=\frac{n_2}{n_1}=\frac{v_1}{v_2}$. For three media, ${}_1{n}_2\times {}_2{n}_3\times {}_3{n}_1=1$. Looking straight down into a liquid, apparent depth = real depth / n. When light goes from a denser to a rarer medium, the refracted ray bends away from the normal; at the critical angle C the angle of refraction is 90°: $\sin C=\frac{n_2}{n_1}$ (= 1/n for glass to air). For i > C there is no refracted ray: total internal reflection (TIR). Conditions: light travels from denser to rarer, and i > C. Uses: total-reflecting prisms (a 45°-90°-45° prism turns a ray through 90° or 180°, as in periscopes and binoculars, because i = 45° exceeds C = 42° for glass), the sparkle of diamond (C ≈ 24°), mirages and optical fibres, where light is guided along a high-index core by repeated TIR at the core-cladding boundary.

Worked example

Find the critical angle for glass of n = 1.5 in air. An optical fibre has a core of n = 1.50 and a cladding of n = 1.45; find the critical angle at the core-cladding boundary.

Try it first, then show the solutionSolution

1. Glass-air: $\sin C=\frac{1}{1.5}=0.667$, C = 41.8°
2. Core-cladding: $\sin C=\frac{1.45}{1.50}=0.967$, C = 75.2°
3. So rays inside the core must hit the boundary at more than 75.2° to the normal (nearly grazing) to be guided.

Answer: 41.8°; 75.2°

Refraction through a prism: minimum deviation

For a prism of angle A, a ray refracted at the two faces with angles $i_1,r_1$ and $r_2,e$ satisfies $r_1+r_2=A$ and deviation $\delta =i_1+e-A$. The i-δ graph falls to a minimum ${\delta }_m$ and rises again. At minimum deviation the ray passes symmetrically: $i_1=e=i$, $r_1=r_2=r=A/2$, and inside an equilateral prism the ray is parallel to the base. Then $i=\frac{A+{\delta }_m}{2}$ and Snell's law gives $n=\frac{\sin \frac{A+{\delta }_m}{2}}{\sin \frac{A}{2}}$. For a thin prism (small A), $\delta =(n-1)A$, independent of i.

Worked example

A 60° prism gives a minimum deviation of 38°. Find its refractive index. What deviation would a thin 5° prism of n = 1.5 produce?

Try it first, then show the solutionSolution

1. $n=\frac{\sin \frac{60^{\circ }+38^{\circ }}{2}}{\sin 30^{\circ }}=\frac{\sin 49^{\circ }}{0.5}=\frac{0.7547}{0.5}=1.51$
2. Thin prism: $\delta =(n-1)A=0.5\times 5^{\circ }=2.5^{\circ }$

Answer: n ≈ 1.51; 2.5°

Dispersion, dispersive power and the rainbow

Refractive index depends on wavelength: violet light (shorter λ) travels more slowly in glass and has larger n, so it is deviated most; red least. Splitting white light into colours is dispersion. For a thin prism, angular dispersion $={\delta }_v-{\delta }_r=(n_v-n_r)A$ and dispersive power $\omega =\frac{{\delta }_v-{\delta }_r}{{\delta }_y}=\frac{n_v-n_r}{n_y-1}$, which depends only on the material. A larger deviation does not mean a larger dispersive power. Rainbow: sunlight entering spherical raindrops is refracted (and dispersed), internally reflected and refracted out. Primary rainbow: one internal reflection, red on the outside (about 42°), violet inside (about 40°). Secondary rainbow: two internal reflections, fainter, colours reversed (red inside), at about 50°-53°. The Sun must be behind the observer.

Worked example

For crown glass, $n_v=1.523$, $n_r=1.513$, $n_y=1.517$. Find its dispersive power and the angular dispersion produced by a thin 6° prism of it.

Try it first, then show the solutionSolution

1. $\omega =\frac{1.523-1.513}{1.517-1}=\frac{0.010}{0.517}=0.0193$
2. Angular dispersion $=(n_v-n_r)A=0.010\times 6^{\circ }=0.06^{\circ }$

Answer: ω ≈ 0.019; 0.06°

Refraction at a single spherical surface

Consider a convex surface (radius R, centre C) separating a rarer medium $n_1$ (object side) from a denser medium $n_2$, forming a real image. For a paraxial ray from object O striking the surface at N, with angles measured from the principal axis, the geometry gives $i=\alpha +\gamma$ and $r=\gamma -\beta$, and Snell's law for small angles, $n_{1}i=n_{2}r$. Substituting $\alpha \approx \frac{h}{-u}$, $\beta \approx \frac{h}{v}$, $\gamma \approx \frac{h}{R}$ and simplifying: $\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R}$. This single result leads to the lens maker's formula and also gives apparent positions of objects seen through curved surfaces.

Worked example

A point object in air is 30 cm in front of a convex glass surface of radius 10 cm (n = 1.5). Where is the image formed?

Try it first, then show the solutionSolution

1. $n_1=1$, $n_2=1.5$, u = −30 cm, R = +10 cm
2. $\frac{1.5}{v}-\frac{1}{-30}=\frac{0.5}{10}\Rightarrow \frac{1.5}{v}=0.05-0.0333=0.0167$
3. v = +90 cm (inside the glass, real)

Answer: 90 cm from the surface, inside the glass (real image)

Lens maker's formula and lenses in a liquid

Apply the spherical-surface formula to the two surfaces of a thin lens of index $n_2$ in a medium $n_1$. The first surface forms an image at $v_1$, which acts as the object for the second; adding the two equations eliminates $v_1$: $\frac{1}{v}-\frac{1}{u}=\left(\frac{n_2}{n_1}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$. For an object at infinity, v = f: $\frac{1}{f}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$, where $n=n_2/n_1$. Biconvex: $R_1>0$, $R_2<0$. Plano-convex: $R_2=\infty$. In a liquid the relative index $n_2/n_1$ falls, so f increases; if the liquid's index equals the lens's, the lens has no effect; if the liquid is denser, a convex lens diverges.

Worked example

A biconvex glass lens (n = 1.5) has radii 20 cm each. Find its focal length in air and in water (n = 4/3).

Try it first, then show the solutionSolution

1. Air: $\frac{1}{f}=0.5\left(\frac{1}{20}+\frac{1}{20}\right)=0.05$, f = 20 cm
2. Water: $n=\frac{1.5}{4/3}=1.125$; $\frac{1}{f^{\prime }}=0.125\times 0.1=0.0125$, f' = 80 cm

Answer: 20 cm in air; 80 cm in water (four times longer)

Lens formula, power and lenses in contact

Lens formula (New Cartesian): $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$, derived for a biconvex lens from the ray diagram (similar triangles) and extended to other lenses. Magnification $m=\frac{v}{u}$. Power $P=\frac{1}{f}$ (f in metres), unit dioptre (D); convex lenses have positive power. For two thin lenses in contact, the image by the first acts as object for the second; adding the lens equations gives $\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}$, so $P=P_1+P_2$. Lens-mirror combinations: trace the light step by step; each image is the object for the next element, taking care with direction after reflection.

Worked example

A convex lens of focal length 20 cm forms an image of an object placed 30 cm from it. Find the image position and magnification. If it is put in contact with a lens of power −2.5 D, what is the focal length of the combination?

Try it first, then show the solutionSolution

1. $\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{20}-\frac{1}{30}=\frac{1}{60}$: v = +60 cm; $m=\frac{v}{u}=\frac{60}{-30}=-2$
2. $P_1=\frac{100}{20}=+5D$; combined $P=5-2.5=+2.5D$; $F=\frac{100}{2.5}=40cm$ (converging)

Answer: Image 60 cm behind the lens, real, inverted, twice the size; combination f = +40 cm

Microscopes and telescopes

Simple microscope (convex lens, focal length f, object inside F): image virtual, erect, magnified. $M=1+\frac{D}{f}$ with the image at the near point D (25 cm); $M=\frac{D}{f}$ with the image at infinity. Compound microscope: objective (short $f_o$, small aperture) forms a real, magnified image just inside the eyepiece's focus; the eyepiece acts as a simple microscope. $M=m_o\times m_e=\frac{v_o}{u_o}\left(1+\frac{D}{f_e}\right)$ for the final image at D; for the image at infinity only the expression $M\approx \frac{L}{f_o}\frac{D}{f_e}$ is needed. Resolving power of a microscope $=\frac{2n\sin \theta }{1.22\lambda }$: improved by shorter λ or oil immersion. Refracting telescope: objective of long $f_o$ and large aperture, eyepiece of short $f_e$. Normal adjustment (image at infinity): $M=\frac{f_o}{f_e}$, length $f_o+f_e$; image at D: $M=\frac{f_o}{f_e}\left(1+\frac{f_e}{D}\right)$. Reflecting telescope (concave mirror objective): no chromatic aberration, a parabolic mirror removes spherical aberration, a large mirror is lighter and easier to support than a lens of the same aperture, and gathers more light.

Worked example

A telescope has an objective of focal length 100 cm and an eyepiece of 5 cm. Find its magnifying power and length in normal adjustment, and its magnifying power when the final image is at 25 cm.

Try it first, then show the solutionSolution

1. Normal adjustment: $M=\frac{f_o}{f_e}=\frac{100}{5}=20$; length $=100+5=105cm$
2. Image at D: $M=\frac{f_o}{f_e}\left(1+\frac{f_e}{D}\right)=20\times 1.2=24$

Answer: 20 and 105 cm; 24 with the image at D

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

• Microscopes and Telescopes: Magnifying Power
• Refraction Through a Prism and Dispersion
• Lens Formula, Lens Maker's Formula and Power

Ray Optics and Optical Instruments formula sheet

Mirror formula

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f},f=\frac{R}{2},m=-\frac{v}{u}$

New Cartesian: concave f < 0, convex f > 0.

Snell's law

$n_1\sin i=n_2\sin r,{}_1{n}_2=\frac{v_1}{v_2}$

${}_1{n}_2\times {}_2{n}_3\times {}_3{n}_1=1$.

Apparent depth

$apparentdepth=\frac{realdepth}{n}$

Viewing along the normal.

Critical angle

$\sin C=\frac{n_2}{n_1}=\frac{1}{n}$

TIR needs dense to rare and i > C.

Prism

$r_1+r_2=A,\delta =i+e-A,n=\frac{\sin \frac{A+{\delta }_m}{2}}{\sin \frac{A}{2}}$

At ${\delta }_m$: i = e, $r=A/2$.

Thin prism

$\delta =(n-1)A$

Small A only.

Dispersion

${\delta }_v-{\delta }_r=(n_v-n_r)A,\omega =\frac{n_v-n_r}{n_y-1}$

ω depends only on the material.

Single spherical surface

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

Light from medium $n_1$ to $n_2$.

Lens maker's formula

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

In air $n_1=1$.

Lens formula and magnification

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

Convex f > 0, concave f < 0.

Power and combination

$P=\frac{1}{f(m)},P=P_1+P_2,\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}$

Thin lenses in contact.

Simple microscope

$M=1+\frac{D}{f}(imageatD);M=\frac{D}{f}(imageat\infty )$

D = 25 cm.

Compound microscope

$M=\frac{v_o}{u_o}\left(1+\frac{D}{f_e}\right)(atD);M\approx \frac{L}{f_o}\cdot \frac{D}{f_e}(at\infty )$

Magnitudes; the final image is inverted.

Refracting telescope

$M=\frac{f_o}{f_e}(normal);M=\frac{f_o}{f_e}\left(1+\frac{f_e}{D}\right)(atD)$

Length in normal adjustment $f_o+f_e$.

Resolving power of a microscope

$RP=\frac{2n\sin \theta }{1.22\lambda }$

n sin θ: numerical aperture.

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.

• VIBGYOR: violet bends most

  Violet has the largest n, so it deviates most: 'Violet Is Bent Greatly'.

  Use it when: Dispersion MCQs.

• Mirror plus, lens minus

  Mirror: 1/v + 1/u. Lens: 1/v − 1/u. Both equal 1/f.

  Use it when: Not mixing the two formulas.

• Mirror doesn't care about water

  Reflection does not depend on the medium, so a mirror's f is the same in water; a lens's f grows in water.

  Use it when: Assertion-reason on lenses and mirrors in liquids.

• Grazing means C

  Grazing emergence: $r_2=C$; then $r_1=A-C$ and $\sin i=n\sin r_1$.

  Use it when: Grazing-emergence numericals.

• Telescope: big over small

  Telescope M = big f / small f (objective over eyepiece). Microscope: both short, objective shorter.

  Use it when: Instrument MCQs.

• Primary: Red Rides on top

  Primary rainbow red outside (top); secondary is the mirror image (red inside).

  Use it when: Rainbow difference questions.

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 5-mark question split into 1 + 1 + 2 + 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: Incomplete prism diagram and missing the minimum-deviation condition

  CISCE examiners' comments, 2025: in the prism derivation, ray diagrams were incomplete, arrows were missing, i, r, e and δm were not marked, steps were skipped and the minimum-deviation condition was not stated.

  Wrong working

  Jumps straight to $n=\sin \frac{A+{\delta }_m}{2}/\sin \frac{A}{2}$.

  Right working

  Diagram, $r_1+r_2=A$, $\delta =i+e-A$, condition i = e and $r_1=r_2$, then Snell's law.

  ✓Do this: Write each relation on its own line.

  Usually costs: 1-2 marks of 3.

• ✗Mistake: Quoting C = 42° without calculating it

  CISCE examiners' comments, 2024: in the grazing-emergence numerical, candidates quoted the critical angle as 42° without calculation, used the prism formula instead, or rounded intermediate values.

  Wrong working

  C = 42°, so $r_1=18^{\circ }$ (and then minimum-deviation formula used).

  Right working

  $\sin C=1/n$ calculated; $r_1=A-C$; $\sin i=n\sin r_1$; round only at the end.

  ✓Do this: Calculate every angle from the given n.

  Usually costs: 1-2 marks.

• ✗Mistake: Wrong sign or unit with lenses in contact

  CISCE examiners' comments, 2025: for two lenses in contact given in dioptres, candidates used the wrong formula, made sign errors and gave the wrong unit for focal length.

  Wrong working

  $F=\frac{1}{2.0+1.5}$, unit 'D'.

  Right working

  $P=2.0+(-1.5)=0.5D$, F = 2 m (focal length in metres, not dioptres).

  ✓Do this: Keep the sign with each power; F is a length.

  Usually costs: 1 mark.

• ✗Mistake: Misplacing the object in a magnifying glass

  CISCE examiners' comments, 2025: asked where to place an object for a magnifying glass, candidates placed it at F, between F and 2F or beyond F, and said the image is real.

  Wrong working

  Object beyond F; image real and magnified.

  Right working

  Object between the optical centre and F; image virtual, erect and magnified, on the same side as the object.

  ✓Do this: A simple microscope only works with the object inside F.

  Usually costs: 1-2 marks.

• ✗Mistake: Misjudging a lens in water

  CISCE examiners' comments, 2025: in an assertion-reason item on a convex lens immersed in water, candidates did not know how focal length depends on the relative refractive index.

  Wrong working

  n of glass relative to water is greater than relative to air.

  Right working

  ${}_w{n}_g=1.5/1.33=1.125<1.5$, so $(n-1)$ falls and f increases.

  ✓Do this: Always use the relative index $n_{lens}/n_{medium}$ in the lens maker's formula.

  Usually costs: 1 mark.

• ✗Mistake: Giving a commercial reason for mirror telescopes, or confusing power with magnifying power

  CISCE examiners' comments, 2024 and 2025: candidates gave cost or ease of handling as the advantage of a reflecting telescope, and wrote 'power' when magnifying power was meant.

  Wrong working

  Mirrors are cheaper; the power of the telescope is 20 D.

  Right working

  No chromatic aberration (and parabolic mirrors remove spherical aberration; large mirrors are lighter). Magnifying power M = 20 (no unit).

  ✓Do this: Give physical advantages; keep 'magnifying power' and 'power of a lens' apart.

  Usually costs: 1 mark each.

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 glass of refractive index 1.5 is about:
   (a) 30°
   (b) 42°
   (c) 48°
   (d) 60°

   Hint

   sin C = 1/n.

   Answer

   (b) 41.8°

2. Q2 · 1 mark · MCQBoard level

   The power of a convex lens of focal length 25 cm is:
   (a) +0.25 D
   (b) +2.5 D
   (c) +4 D
   (d) −4 D

   Hint

   f in metres.

   Answer

   (c) +4 D

3. Q3 · 1 mark · MCQBoard level

   When a prism is set for minimum deviation:
   (a) the angle of incidence is 90°
   (b) the refracted ray inside is parallel to the base (equilateral prism)
   (c) the emergent ray grazes the second face
   (d) $r_1=2r_2$

   Hint

   Symmetric passage.

   Answer

   (b)

4. Q4 · 1 mark · Very short answerBoard level

   Name the three optical processes that take place inside a raindrop to form a primary rainbow.

   Hint

   In, back, out.

   Answer

   Refraction with dispersion on entry, (total) internal reflection at the back, and refraction on leaving.

5. Q5 · 2 marks · Short answerBoard level

   An object is 25 cm in front of a concave mirror of focal length 15 cm. Find the image distance and magnification.

   Hint

   1/v + 1/u = 1/f with signs.

   Answer

   v = −37.5 cm (real), m = −1.5 (inverted, magnified).

6. Q6 · 2 marks · Short answerHarder board level

   A coin lies at the bottom of a beaker containing water 12 cm deep (n = 4/3). By how much does the coin appear raised when viewed from above?

   Hint

   Apparent depth = real/n.

   Answer

   Apparent depth = 9 cm, so it appears raised by 3 cm.

7. Q7 · 2 marks · Short answerHarder board level

   For crown glass $n_v=1.523$, $n_r=1.513$, $n_y=1.517$. Find the dispersive power, and the mean deviation produced by a thin 6° prism.

   Hint

   ${\delta }_y=(n_y-1)A$.

   Answer

   ω = 0.0193; ${\delta }_y=0.517\times 6^{\circ }=3.1^{\circ }$.

8. Q8 · 2 marks · Short answerHOTS / competency

   A glass lens (n = 1.5) is placed in a liquid and becomes invisible. What is the refractive index of the liquid, and what is the focal length of the lens in it? Explain.

   Hint

   Relative index.

   Answer

   1.5. Relative index = 1, so $(n-1)=0$ and f is infinite: no refraction or reflection at the surfaces, so the lens cannot be seen.

9. Q9 · 3 marks · Short answerHarder board level

   A magnifying glass has f = 5 cm. Find its magnifying power with the image at 25 cm and at infinity, and the object distance for the image at 25 cm.

   Hint

   $M=1+D/f$ and $D/f$.

   Answer

   M = 6 (image at D), 5 (at infinity); object 4.17 cm from the lens (inside F).

10. Q10 · 3 marks · Short answerHarder board level

    An astronomical telescope has $f_o=100$ cm and $f_e=5$ cm. Find the magnifying power and tube length in normal adjustment, and the magnifying power with the final image at 25 cm.

    Hint

    $f_o/f_e$.

    Answer

    20; 105 cm; 24.

11. Q11 · 3 marks · Short answerHarder board level

    Derive the relation $\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R}$ for refraction at a convex surface. Use it for an object in air 30 cm in front of a convex glass surface (R = 10 cm, n = 1.5).

    Hint

    Use i = α + γ and r = γ − β with small angles.

    Answer

    With $n_{1}i=n_{2}r$, $\alpha =h/(-u)$, $\beta =h/v$, $\gamma =h/R$: $n_1(\alpha +\gamma )=n_2(\gamma -\beta )$ gives the result. Numerical: $\frac{1.5}{v}=\frac{0.5}{10}-\frac{1}{30}$, v = +90 cm.

12. Q12 · 5 marks · Long answerHarder board level

    (a) Derive the lens maker's formula. [3]
    (b) A biconvex lens (n = 1.5) has radii 20 cm. Find its focal length and power, and the focal length when placed in contact with a concave lens of focal length 10 cm. [2]

    Hint

    Two refractions; the first image is the second object.

    Answer

    (a) First surface: $\frac{n}{v_1}-\frac{1}{u}=\frac{n-1}{R_1}$; second: $\frac{1}{v}-\frac{n}{v_1}=\frac{1-n}{R_2}$; adding: $\frac{1}{v}-\frac{1}{u}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)=\frac{1}{f}$. (b) f = 20 cm, P = +5 D; combination $\frac{1}{F}=\frac{1}{20}-\frac{1}{10}$, F = −20 cm.

13. Q13 · 2 marks · Short answerHarder board level

    An object is 30 cm in front of a convex lens of focal length 20 cm. A plane mirror is placed 40 cm behind the lens, facing it. Trace the light and find where the final image is formed after the light passes back through the lens.

    Hint

    Lens image → mirror image → lens again.

    Answer

    Lens: v = +60 cm, i.e. 20 cm behind the mirror. The mirror reflects the converging light to form an image 20 cm in front of it, i.e. 20 cm from the lens. That point is at the lens's focus, so after passing back through the lens the light emerges parallel: final image at infinity.

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.Mirror formula and magnification?Show answer

   $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$, $m=-v/u$.

2. 2.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.Prism formula at minimum deviation?Show answer

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

4. 4.Define dispersive power.Show answer

   $\omega =\frac{n_v-n_r}{n_y-1}$: angular dispersion divided by mean deviation.

5. 5.Lens maker's formula?Show answer

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

6. 6.Focal length of lenses in contact?Show answer

   $\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}$; $P=P_1+P_2$.

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

   $f_o/f_e$.

8. 8.One difference between primary and secondary rainbows?Show answer

   Primary: one internal reflection, red outside. Secondary: two reflections, colours reversed, fainter.

9. 9.One advantage of a reflecting telescope?Show answer

   No chromatic aberration (also: less spherical aberration with a parabolic mirror; large mirrors are lighter).

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you derive the mirror formula and use it with signs?
• ☐Can you find the critical angle and explain an optical fibre and a total-reflecting prism?
• ☐Can you derive the prism formula with a complete diagram?
• ☐Can you solve a grazing-emergence problem?
• ☐Can you find dispersive power and describe primary and secondary rainbows?
• ☐Can you derive the spherical-surface and lens maker's formulas?
• ☐Can you find f of a lens in water and of lenses in contact?
• ☐Can you draw and derive the magnifying power of a simple microscope, a compound microscope and a telescope?
• ☐Can you state two advantages of a reflecting telescope?

Ray Optics and Optical Instruments: frequently asked questions