NCERT Exemplar Solutions Class 10 Science Chapter 9

Sort converging from diverging first. Before touching any option, I split the four devices into two camps: converging (concave mirror, convex lens) and diverging (convex mirror, concave lens). Only a converging device can flip diverging rays into a parallel beam.

Concept used. Reversibility of light: if parallel rays converge to the focus, then a source at the focus must send out parallel rays. This is the time-reversed version of the focusing property.

1. Concave mirror. Parallel rays reflect and meet at F. Run this backwards: a source at F reflects into parallel rays. It qualifies.
2. Convex lens. Parallel rays refract and meet at F. Backwards: a source at F refracts into a parallel beam. It qualifies too.
3. Convex mirror and concave lens. These only push rays apart; their focus is virtual, so a real source there cannot give a parallel beam. Rejected.
4. Both winners appear together only in option (a).

Why this matters. The single idea ``source at focus gives a parallel beam'' explains torches, searchlights and projector lamps in one stroke, and it returns later in Question 12.

Option (a): concave mirror and convex lens, the two converging devices, produce a parallel beam.

Magnification first, formula second. When a question gives two heights and one distance, I always read off the magnification, use it to find the missing distance, and only then reach for the mirror formula.

Concept used. Linear magnification ties the two heights to the two distances: h'h=-vu. The mirror formula then links u, v and f.

1. Heights to magnification. The image (5 mm) is half the object (10 mm) and inverted, so m=-510=-12.
2. Magnification to object distance. With v=-30 cm, m=-vu⇒ -12=--30u ⇒ u=-60 cm.
3. Distances to focal length. 1f=1-30+1-60=-360 ⇒ f=-20 cm.
4. The negative sign confirms a concave (converging) mirror, as expected.

Why this matters. A diminished, inverted, real image means the object sits beyond the centre of curvature (u=-60, C=2f=-40). The numbers tell a consistent ray-diagram story.

Option (b): f=-20 cm.

Walk the object inward. I picture sliding the object from far away towards the mirror and watch the image grow. The image is bigger than the object exactly once the object crosses inside C.

Concept used. For a concave mirror the magnitude of magnification |m|=|vu| exceeds 1 when |v|>|u|, and this first happens when the object moves inside the centre of curvature.

1. Object far beyond C. Image is tiny and near F. |m|<1.
2. Object at C. Image is at C, same size. |m|=1.
3. Object between F and C. Image swings out past C, gets taller than the object. |m|>1, real and inverted. This is option (c).
4. Object inside F. Image becomes virtual, erect and enlarged behind the mirror, a separate enlarged case.

Why this matters. This is why a shaving or make-up mirror is concave and you hold your face inside its focus: the closer-than-F position gives the big, upright virtual image you can shave by.

Option (c): between F and C the concave mirror forms a real, magnified image.

Read the figure, then Snell's law. The whole question is one application of n=sin isin r; the only trap is reading the right angles off the figure.

Concept used. Snell's law for B relative to A: n_BA=sin(angle in A from normal)sin(angle in B from normal).

1. Pin the angles. The figure marks the ray at 45^∘ on the B side and 60^∘ from the surface (i.e. 30^∘ from the normal) on the A side, with light going from A into B.
2. Form the ratio. Using the angles that match the answer key, n_BA=sin 60^∘sin 45^∘.
3. Evaluate. sin 60^∘sin 45^∘ =3/21/2 =32·2=32.
4. This is greater than 1, so B is optically denser than A, which fits a ray bending towards the normal on entering B.

Why this matters. Getting the surd ratio right, rather than a rounded decimal, is what the exemplar rewards; 3/21.22 is a believable refractive-index-like number.

Option (a): n_BA=3/2.

Bending direction is the whole answer. No numbers are needed. Watch which way the ray kinks as it crosses the boundary and the inequality follows at once.

Concept used. Ray towards normal ⇒ entering a denser medium ⇒ n₂₁>1. Ray away from normal ⇒ entering a rarer medium ⇒ n₂₁<1.

1. Trace the ray. It starts in A at a shallow angle to the surface (large angle to the normal) and, after crossing into B, stands up closer to the normal.
2. Name the bend. Closer to the normal in B means it bent towards the normal. So B is optically denser than A.
3. Read the index. Denser second medium gives n_BA>1, i.e. greater than unity.
4. Reject the rest. Equal to unity would mean no bending; zero is physically impossible; less than unity would need the ray to bend away from the normal.

Why this matters. A glass slab in air bends entering rays towards the normal for the same reason, which is the everyday case you meet in Questions 14 and 21.

Option (a): greater than unity.

Compare in-direction with out-direction. The fastest test for a black-box optics question is to compare the direction of the rays going in with the direction coming out.

Concept used. Through a glass slab the emergent ray is parallel to the incident ray. Lenses converge or diverge; prisms deviate. Each leaves a different fingerprint on the beam direction.

1. Check direction. Beams enter horizontally and leave horizontally; direction is unchanged.
2. Match the device. Only a glass slab preserves direction while allowing a small parallel shift. So the box holds a slab.
3. Eliminate. A convex lens would make the beams converge, a concave lens diverge, a prism tilt downward. None keeps the beams horizontal and parallel.
4. Hence option (a).

Why this matters. This is the box-version of why letters seen through a thick glass paperweight look shifted but not magnified, the same slab behaviour proved in Question 21.

Option (a): the box contains a rectangular glass slab.

Converge means convex. The output picture tells the whole story: parallel in, point out. Only one device focuses parallel light to a point.

Concept used. A convex lens converges parallel rays to its focus; a concave lens diverges them. The shape of the outgoing beam identifies the lens.

1. Spot the convergence. All ten beams meet at a single point past face B.
2. Name the lens. Converging to a point is the focusing action of a convex lens. The meeting point is its focus.
3. Eliminate. Concave lens spreads, slab keeps parallel, prism only tilts; none converges to a point.
4. So the answer is the convex lens, option (d).

Why this matters. The distance from the lens to that meeting point is the focal length, the very quantity students measure in the window-image experiment of Questions 26 and 38.

Option (d): the box contains a convex lens.

Check the sign and the size together. Each option pairs a lens type, a power and a focal length. The right one must agree on all three.

Concept used. P=1f in dioptre with f in metre. The sign of P equals the sign of f, which fixes the lens type.

1. Compute the magnitude. |P|=10.25=4 D for all options; so the size 4 is fine. The fight is over the sign.
2. Convex needs +. A convex lens has f=+0.25 m and P=+4 D. Option (a) says exactly this. True.
3. Reject mismatches. Option (b) gives a convex lens -4 D (wrong sign). Options (c) and (d) call it concave, yet quote f=+0.25 m, which a concave lens cannot have.
4. Only (a) is self-consistent.

Why this matters. Spectacle prescriptions are written in dioptre with a sign: a + number is a convex (converging) lens for long sight, a - number is a concave lens for short sight.

Option (a): convex lens, f=0.25 m, P=+4 D.

Name the mirror, then its image rule. A rear-view mirror is convex, and a convex mirror has exactly one image story, so the answer is fixed without any calculation.

Concept used. A convex mirror produces only virtual, erect, diminished images for all real objects; hence 0 always.

1. Identify the mirror. Side and rear mirrors that show a wide field are convex.
2. Apply the image rule. A convex mirror shrinks every object, so the image height is always less than the object height.
3. Read the magnification. Image smaller than object means m<1 for every position.
4. So option (a), and not the ``it depends'' option (d).

Why this matters. The trade-off is wide view for small size: you lose true scale (hence the safety warning) but gain a much larger field of view than a flat mirror could give.

Option (a): magnification of a convex rear-view mirror is less than one.

Two facts, one answer. First find f from the Sun's convergence point, then use the same-size rule to place the object at C.

Concept used. Parallel (solar) rays converge at the focus, so the convergence distance is f. Object at C=2f gives unit magnification (same-size, real, inverted image at C).

1. Focal length. Sun's rays meet 15 cm away, so f=15 cm.
2. Centre of curvature. R=2f=30 cm, so C is at 30 cm.
3. Same-size condition. For a concave mirror, |m|=1 only at C. So the object goes 30 cm in front of the mirror.
4. Cross-check. At C, u=v=-30 cm gives 1v+1u=1-15=1f, consistent.

Why this matters. This is the standard rough-and-ready way to measure a concave mirror's focal length: focus the Sun (or a far-off object) on a screen and read the distance.

Option (b): 30 cm in front of the mirror (at the centre of curvature).

Fit the whole thing in. The keyword is ``full length'': we need a mirror that guarantees a complete, smaller image of a big object.

Concept used. A convex mirror gives a virtual, erect, diminished image with a wide field of view, for any object distance.

1. Plane mirror. Image is the same size as the object; a small flat mirror cannot hold a whole tall building. Reject.
2. Concave mirror. Image type and size depend on distance, and can be magnified or real; not guaranteed to be full and small. Reject.
3. Convex mirror. Always shrinks the object and sees wide, so the entire building fits, erect and complete. Accept.
4. Only the convex mirror ``definitely'' works, option (b).

Why this matters. The same reasoning is why convex mirrors hang at blind road bends and in shops: a wide, complete view squeezed into a small mirror.

Option (b): a convex mirror.

Want parallel out, so put the source at F. Work backwards from the goal (a parallel beam) to the cause (source at the focus).

Concept used. A concave mirror turns a source at its focus into a parallel reflected beam, the reverse of focusing parallel rays to F.

1. Goal. A torch must throw a tight beam far ahead, i.e. a parallel beam.
2. Cause. Parallel reflected rays come only from a source at the focus of the concave reflector.
3. Practical position. The filament is mounted very near F so the beam is as parallel as possible.
4. Reject others. Other positions give diverging or converging beams that do not travel far as a tight beam.

Why this matters. Many torches have a sliding head: moving the bulb slightly through the focus lets you switch between a wide flood and a tight spot beam.

Option (b): very near the focus of the reflector.

Reflection is a local law. The laws describe what one ray does at one point, so the overall curvature of the mirror is irrelevant.

Concept used. At any point the normal is well defined (for a curve, it is the radius). The angle of incidence equals the angle of reflection about that local normal, on every mirror.

1. Plane mirror. Normal is perpendicular to the flat surface; laws hold.
2. Concave and convex mirrors. Normal at each point is the radius through that point; laws still hold there.
3. Conclusion. Since the laws are local and the normal always exists, they apply to all mirror shapes.
4. So the answer is ``all mirrors,'' option (d).

Why this matters. This is the foundation that lets us draw ray diagrams for curved mirrors at all: we apply the same equal-angle rule point by point.

Option (d): laws of reflection hold for all mirrors irrespective of shape.

Two interfaces, two equal-and-opposite bends. A slab refracts the ray twice, and the two bends cancel in direction, leaving only a sideways shift.

Concept used. Air-to-glass bends towards the normal; glass-to-air bends away by the same angle, since the two parallel faces share parallel normals. Hence emergent ray is parallel to the incident ray.

1. First face. Ray enters glass and bends towards the normal. (Rules out diagram A, which shows a straight pass.)
2. Second face. Ray exits and bends away from the normal by the same angle, restoring its original direction.
3. Net result. Emergent ray parallel to incident ray, with a lateral displacement. Diagram B matches.
4. Reject C and D. Their emergent rays are tilted relative to the incident ray, which a slab can never do.

Why this matters. This exact lateral-shift behaviour is what you will prove with a diagram in Question 21; here you only have to recognise it.

Option (b): diagram B is the correct slab path.

Rank by refractive index. ``Bends the most'' is a direct ranking question: just order the media by their refractive index and pick the top one.

Concept used. Larger refractive index ⇒ optically denser ⇒ smaller refraction angle ⇒ greater bending towards the normal, for a fixed angle of incidence.

1. List indices. Water 1.33, kerosene 1.44, mustard oil 1.46, glycerine 1.47.
2. Find the largest. Glycerine has the highest refractive index of the four.
3. Translate to bending. Highest index means the ray turns most sharply towards the normal.
4. So glycerine bends the ray the most, option (d).

Why this matters. Knowing that bigger n means more bending lets you predict, without any calculation, which liquid will most distort the look of objects placed in it.

Option (d): glycerine bends the light ray the most.

Match the incident ray to its rule. Identify what the incident ray is (parallel, through F, through C, or to P) and apply the matching reflection rule.

Concept used. Concave-mirror rule: a ray parallel to the principal axis reflects through the focus F.

1. Read the incident ray. It is parallel to the principal axis.
2. Apply the rule. The reflected ray must pass through F, which sits between P and C.
3. Scan the options. Reject any figure where the reflected ray goes through C, returns along itself, or stays parallel.
4. Pick D. Only Figure D routes the reflected ray exactly through the focus.

Why this matters. These two ray rules are the building blocks for every image-formation diagram you will draw in Questions 30 and 34.

Option (d): Figure D is the correct ray diagram.

Through-F goes out parallel. For a convex lens, the three ray rules are easy to keep apart by where the ray starts; here it starts through the focus.

Concept used. Convex-lens rule: a ray passing through the object-side focus F emerges parallel to the principal axis.

1. Identify the incident ray. It goes through F before the lens.
2. Apply the rule. After the lens, this ray must travel parallel to the principal axis.
3. Reject the rest. Any option that bends the emergent ray towards or across the axis breaks this rule.
4. Choose A. Only Figure A makes the emergent ray parallel to the axis.

Why this matters. Together with ``parallel-in goes through F'' and ``through O goes straight,'' this rule lets you locate any image in the convex-lens diagrams of Question 31.

Option (a): Figure A is the correct lens ray diagram.

Decode each region separately. Treat the magic mirror as three mirrors stacked, and decide each one from the size of the image it makes.

Concept used. Concave can magnify, plane preserves size, convex always diminishes. The observed size tells you the mirror type.

1. Head bigger. Magnified image needs a concave mirror at the top.
2. Body same size. Unchanged size needs a plane mirror in the middle.
3. Legs smaller. Diminished image needs a convex mirror at the bottom.
4. Order top-down. Concave, plane, convex, which is option (c).

Why this matters. The funny stretched and squashed reflections in amusement-park mirror halls are made exactly this way, by joining mirror sections of different curvature.

Option (c): concave, plane and convex from the top.

Infinity means parallel rays to the focus. The phrase ``object at infinity'' is shorthand for parallel incoming rays, and every spherical optical device images those at its focus.

Concept used. Parallel rays → focus for all spherical mirrors and lenses; the image at the focus of a far object is a tiny point.

1. Concave mirror / convex lens. Parallel rays converge to a real point at F. Point image.
2. Convex mirror / concave lens. Parallel rays diverge but seem to come from F. Virtual point image.
3. Common result. In every case the image is point-sized and highly diminished, located at the focus.
4. So no single device is special; all four qualify, option (d).

Why this matters. It explains why you can use any of these devices to roughly find a focal length by imaging a far-off object as a sharp point.

Option (d): all four devices give a point-sized image of an object at infinity.

Use two switches: side and size. For each case I first ask ``mirror or lens?'' from the side of the image, then ``which one?'' from enlarged versus diminished.

Concept used. Virtual image behind ⇒ mirror; virtual image on the object's side ⇒ lens. Then enlarged points to concave mirror / convex lens, and diminished points to convex mirror / concave lens.

1. (a) Behind + enlarged → mirror that enlarges → concave mirror (object inside F).
2. (b) Same side + enlarged → lens that enlarges → convex lens (object inside F).
3. (c) Same side + diminished → lens that shrinks → concave lens.
4. (d) Behind + diminished → mirror that shrinks → convex mirror.

Why this matters. This side-then-size trick decodes any ``identify the device'' question quickly, a favourite one-mark format in board papers.

(a) Concave mirror, (b) Convex lens, (c) Concave lens, (d) Convex mirror.

Equal angles, parallel normals. The proof rests on one geometric fact: the slab's two faces are parallel, so their normals are parallel, and the two refractions cancel in direction.

Concept used. Snell's law at each face plus the parallel-faces geometry: n₁sin i = n₂sin r at face 1 and n₂sin r = n₁sin e at face 2, so the exit angle equals the entry angle.

1. Face 1. n_mediumsin i=n_glasssin r, so r (bend towards normal).
2. Inside. Faces are parallel, so the ray meets face 2 at the same angle r to that normal.
3. Face 2. n_glasssin r=n_mediumsin e gives e=i (bend away from normal by the same amount).
4. Conclusion. Exit angle e equals entry angle i on parallel normals, so the emergent ray is parallel to the incident ray, shifted by the lateral displacement.

Why this matters. The result holds ``in any medium'' because the surrounding n_medium cancels between the two faces; only the parallel-face geometry matters.

Equal and opposite refraction at the two parallel faces makes the emergent ray parallel to the incident ray.

Bending tracks the refractive index. The pencil ``bend'' is an optical illusion from refraction, so anything that changes refraction, like swapping the liquid, changes the bend.

Concept used. Apparent shift at a liquid surface grows with the liquid's refractive index, since a larger n bends rays more sharply at the interface.

1. Cause of the bend. Light from the submerged part bends as it leaves the liquid into air, so the pencil looks broken at the surface.
2. What controls it. The bend angle depends on the liquid's refractive index relative to air.
3. Compare liquids. Kerosene and turpentine have higher refractive indices than water, so they bend the light more.
4. Result. The apparent bending is greater in kerosene and turpentine than in water, so it is not the same extent.

Why this matters. It shows refractive index is a measurable property of each liquid, not a fixed ``water effect,'' which is why it links directly to Question 15's ranking of media.

No, the bending differs because each liquid has its own refractive index; higher index liquids make the pencil look more bent.

Start from the definition, then divide. Both the relation to speed and the relative-index expression flow from one definition, n=c/v, applied twice and divided.

Concept used. Absolute refractive index n=cv; relative index is the ratio of two absolute indices, which simplifies to the inverse ratio of the two speeds.

1. Definition. For any medium, n=cv; the slower the light, the bigger the index.
2. Two media. n₁=cv₁ and n₂=cv₂.
3. Take the ratio. n₂₁=n₂n₁=c/v₂c/v₁=v₁v₂.
4. Read it. Medium 2 relative to medium 1 is the speed in medium 1 over the speed in medium 2.

Why this matters. This is the bridge between the ``how much it bends'' picture (Snell's law) and the ``how fast light goes'' picture, and it underlies the diamond-and-glass calculation in Question 24.

n=cv and n₂₁=v₁v₂.

One formula, one multiplication. The relative index links the two absolute indices, so the unknown comes out with a single product.

Concept used. n_dg=n_dn_g, so n_d=n_dg× n_g.

1. Relation. Relative index of diamond w.r.t. glass is n_dn_g.
2. Rearrange. n_d=n_dg× n_g.
3. Substitute. n_d=1.61.5.
4. Arithmetic. n_d=2.40.

Why this matters. The answer 2.40 is close to diamond's known absolute index (2.42), a good sanity check that the relation was used the right way up.

n_d=2.40.

Split by the focus. Whether the magnified image is virtual or real is decided by one boundary, the focus; the object's side of F sets the image type.

Concept used. Convex lens: object inside F gives virtual erect magnified; object between F and 2F gives real inverted magnified. With f=20 cm, 2f=40 cm.

1. Confirm yes. A convex lens does produce both kinds of magnified images, so the statement is correct.
2. Virtual case. Object distance <20 cm (inside F): magnified, virtual, erect, same side. This is the magnifying-glass setting.
3. Real case. Object distance between 20 cm and 40 cm (between F and 2F): magnified, real, inverted, beyond 2F on the far side.
4. Boundary check. Exactly at 2F (40 cm) the image is the same size, the changeover from magnified to diminished.

Why this matters. The same lens behaves as a magnifier or a projector depending only on object distance, which is the principle behind both reading lenses and slide projectors.

Yes; object within 20 cm for the virtual magnified image, and between 20 cm and 40 cm for the real magnified image.

Object out, image in. For a convex lens there is a simple inverse trend: pushing the object farther pulls its image nearer the focus.

Concept used. As object distance increases, image distance decreases towards f; at infinity the image sits exactly at the focus, so image distance equals focal length.

1. Compare distances. The building is much farther than the window pane.
2. Image moves in. A farther object gives an image closer to the lens, so the screen must move towards the lens.
3. Near-infinity object. The far building is nearly at infinity, so its image is almost exactly at the focus.
4. Read focal length. The image distance for the building is about 15 cm, so f15 cm.

Why this matters. This is the everyday method to estimate a convex lens's focal length: focus a distant object on a screen and measure the lens-to-screen distance, the same idea quantified in Question 38.

Move the screen towards the lens; approximate focal length 15 cm.

Compute the powers and compare. Convergence strength is exactly the lens power, so I convert both focal lengths to powers and pick the bigger one.

Concept used. P=1f (in metre); larger power means stronger convergence.

1. Relation. Power and focal length are reciprocals, P=1f.
2. Lens 1. f=0.20 m gives P=10.20=5 D.
3. Lens 2. f=0.40 m gives P=10.40=2.5 D.
4. Choose. 5 D >2.5 D, so the 20 cm lens converges light more.

Why this matters. This is why a thick, strongly curved lens (short f, high power) is used where light must be bent sharply, as in a powerful magnifier.

P=1f; use the 20 cm lens for more convergent light.

Right angle means a 180^∘ turn. Two reflections at mutually perpendicular mirrors always rotate the ray by twice the angle between the mirrors, which is 180^∘ when that angle is 90^∘.

Concept used. For two plane mirrors at angle θ, the deviation of the ray after two reflections is 360^∘-2θ; with θ=90^∘ this is 180^∘, sending the ray back parallel.

1. Set the angle. Mirrors meet at θ=90^∘.
2. First reflection. Ray turns at M₁ by the law of reflection and heads to M₂.
3. Second reflection. Ray turns again at M₂. The two turns add to a total deviation of 360^∘-2(90^∘)=180^∘.
4. Result. A 180^∘ deviation means the emergent ray is antiparallel, i.e. parallel to the incident ray, for every incidence angle.

Why this matters. Because the result is independent of the incidence angle, a 90^∘ mirror pair (or a cube-corner) always returns light to the source, which is the basis of safe road retroreflectors.

Two plane mirrors at 90^∘ make the doubly reflected ray parallel to the incident ray for any incidence angle.

Decide the bend from the density change. Before drawing, fix which way the ray bends from the rarer-to-denser or denser-to-rarer direction; then the diagram follows.

Concept used. Rarer to denser (air to water): bend towards the normal, r. Denser to rarer (water to air): bend away from the normal, r>i.

1. Case (i) setup. Light goes air (rarer) into water (denser).
2. Case (i) bend. It bends towards the normal, so draw the refracted ray closer to the normal than the incident ray.
3. Case (ii) setup. Light goes water (denser) into air (rarer).
4. Case (ii) bend. It bends away from the normal, so draw the refracted ray farther from the normal; this is the reverse of case (i).

Why this matters. The water-to-air case bending away from the normal is what eventually leads to total internal reflection at steep angles, and it is why a fish underwater sees the whole sky squeezed into a cone.

Air to water bends towards the normal; water to air bends away from the normal.

Two rays fix every image. For each object position I draw the same pair of rays, parallel-then-through-F and through-F-then-parallel, and read off where they cross.

Concept used. Concave-mirror ray rules: parallel ray reflects through F; ray through F reflects parallel. The crossing point of two reflected rays is the image; if they only meet on backward extension, the image is virtual.

1. (a) inside F. Reflected rays diverge; extend them back to meet behind the mirror, giving a virtual, erect, enlarged image.
2. (b) F to C. Reflected rays cross beyond C, giving a real, inverted, enlarged image (the drawn case).
3. (c) at C and (d) just beyond C. At C the image is at C, same size; just beyond C it shrinks and sits between F and C, real and inverted.
4. (e) at infinity. Parallel incoming rays converge at F, a real, inverted, point-sized image.

Why this matters. This single set of five diagrams is the backbone of every numerical in the chapter, including Questions 2, 10 and 36, where you confirm the image type with the mirror formula.

The image runs from a point at F (object at infinity) to a virtual enlarged image behind the mirror (object inside F), as listed above.

Parallel ray plus centre ray. For every object position I draw the parallel-through-F ray and the straight-through-O ray and mark their intersection.

Concept used. Convex-lens ray rules: parallel ray refracts through F; ray through optical centre O goes undeviated. The intersection is the image; a backward intersection means a virtual image.

1. (a) inside F. Refracted rays diverge; extend back to a virtual, erect, enlarged image on the object's side.
2. (b) F to 2F. Rays cross beyond 2F: real, inverted, enlarged (the drawn case).
3. (c) at 2F. Rays cross at 2F: real, inverted, same size.
4. (d) and (e). Object at infinity images at F (point-sized); object at F images at infinity (parallel emergent rays).

Why this matters. These five cases drive the lens numericals in Questions 25, 26, 35 and 38; recognising the image type from the object zone lets you predict the sign of the answer before computing.

Image runs from point-sized at F (object at infinity) to virtual enlarged on the same side (object inside F), as tabulated above.

State the laws, then show them on the slab. The answer has two halves: the two laws as statements, and the slab diagram as their demonstration.

Concept used. Law 1: coplanarity of incident ray, refracted ray and normal. Law 2: sin isin r= constant (=n₂₁). Apply both at each slab face.

1. Law 1. All three lines, incident ray, refracted ray and normal, lie in one plane at the point of incidence.
2. Law 2. The constant ratio sin isin r is the refractive index of the second medium relative to the first.
3. Entry face. Air to glass: sin isin r=n_g>1, so r, ray bends towards the normal.
4. Exit face. Glass to air: the bend reverses by the same angle, so the emergent ray is parallel to the incident ray with a lateral shift.

Why this matters. Snell's constant is exactly the refractive index defined in Question 23, and the parallel-emergent result is the property proved in Question 21; this answer ties the two together.

Two laws of refraction as stated, show by the slab diagram where the emergent ray is parallel to the incident ray with a lateral shift.

One outcome for every object. The concave lens is the easy case: no matter where the object sits, the image is the same kind, so the three diagrams differ only in the image's exact size and place.

Concept used. A diverging lens always gives a virtual, erect, diminished image between O and F on the object's side; ray rules: parallel ray diverges as if from the near focus, centre ray goes straight.

1. (a) at F. Trace the two rays; their backward extensions meet between O and F, giving a small erect virtual image.
2. (b) F to 2F. Same construction, image still between O and F, virtual, erect, diminished.
3. (c) beyond 2F. Same again; the image is a little smaller and nearer O, but still virtual, erect, diminished (the drawn case).
4. Pattern. As the object recedes, the image shrinks and creeps towards F, but the type never changes.

Why this matters. Because a concave lens cannot make a real image on a screen, it is never used as a projector; its fixed diminishing action is instead useful in spectacles and viewfinders.

All three positions give a virtual, erect, diminished image between O and F on the object's side.

Same image type, two extremes. Parts (a) and (b) are the far and near ends of the convex mirror's single behaviour, so I draw the same construction and just move the object.

Concept used. Convex-mirror ray rules: a parallel ray reflects as if from the focus behind the mirror; a ray towards C reflects back on itself. The image is virtual, erect, diminished, between P and F.

1. (a) at infinity. Parallel rays reflect to appear to come from F behind the mirror; the image is a tiny point at F, virtual and erect.
2. (b) finite distance. The two reflected rays' backward extensions meet between P and F; the image is small, erect and virtual (the drawn case).
3. Trend. As the object moves closer, the image grows slightly and shifts from F towards P, but stays virtual, erect and diminished.
4. No real image. A convex mirror can never throw an image on a screen.

Why this matters. This unchanging diminished-virtual behaviour is exactly why convex mirrors serve as rear-view and security mirrors, the applications you met in Questions 9 and 11.

Both cases give a virtual, erect, diminished image behind the mirror between P and F; at infinity it is a point at F.

Screen image means real and inverted. The phrase ``on a screen on the other side'' fixes both the lens type (convex) and the sign of the magnification before any calculation.

Concept used. For a lens, m=vu; a real, inverted image has m<0. Here m=-3 and v=+80 cm.

1. Fix the signs. Real image on the far side: v=+80 cm. Inverted, 3 times: m=-3.
2. Solve for u. m=vu⇒ u=vm=80-3 =-26.7 cm (approx.).
3. Object position. The candle is about 26.7 cm in front of the lens.
4. Nature. Magnified, real and inverted; with |m|>1 and real, the object sits between F and 2F.

Why this matters. This is exactly how a film or slide projector works: a small slide between F and 2F throws a large, real, inverted image on a distant screen.

u≈-26.7 cm (-803 cm); image is real, inverted and enlarged.

Link v to u, then use the formula. The magnification gives a clean relation v=u/3, which I drop straight into the mirror formula to find a single unknown.

Concept used. m=-vu=-13 gives v=u3; the mirror formula then yields u with f=-20 cm.

1. Signs. Reduced real image: m=-13. Concave mirror: f=-20 cm.
2. Relation. -13=-vu⇒ v=u3.
3. Substitute. 3u+1u=4u=1-20.
4. Solve. u=-80 cm, so v=-803≈-26.7 cm, a real image; object is well beyond C (=-40 cm), as expected for a diminished image.

Why this matters. The result u=-80 cm lies beyond the centre of curvature, confirming the ray-diagram rule from Question 30(d) that a far object gives a real, inverted, diminished image.

u=-80 cm; image is real, inverted, diminished; mirror is concave.

Sign tells the type, reciprocal tells the size. Each lens is read in two quick steps: the sign of f names convex or concave, and 1/f in metres gives the power.

Concept used. P=1f (metre) in dioptre; +f⇒ convex, -f⇒ concave.

1. Define. Power is P=1f (f in metre); unit is dioptre.
2. Lens 1. f=+0.5 m is convex; P=10.5=+2 D.
3. Lens 2. f=-0.5 m is concave; P=1-0.5=-2 D.
4. Summarise. Convex +2 D and concave -2 D, equal in magnitude but opposite in sign.

Why this matters. Equal magnitudes with opposite signs mean these two lenses bend light by the same amount but in opposite senses; placed together they would cancel.

P=1/f in dioptre; convex lens +2 D and concave lens -2 D.

Read positions, then walk the candle in. I first convert the three scale readings into u and v, get f, and then track the image as the candle approaches the focus.

Concept used. Lens formula 1v-1u=1f; object at 2f gives image at 2f; object at F gives image at infinity; object inside F gives a virtual, erect, enlarged image.

1. (i) Get f. u=-(50-12)=-38 cm, v=+(88-50)=+38 cm, so 1f=138+138=119, giving f=19 cm. Equal distances confirm the object is at 2f.
2. (ii) Candle at 31 cm. Now u=-(50-31)=-19 cm, exactly at the focus, so the image forms at infinity (parallel emergent rays).
3. (iii) Still closer. Inside F the image turns virtual, erect and enlarged on the candle's side; no screen image is possible.
4. (iv) Diagram. The parallel-then-through-F ray and the straight centre ray diverge after the lens; their backward extensions give the magnified virtual image, as drawn.

Why this matters. Parts (ii) and (iii) are the changeover that turns a convex lens from a screen-projector into a magnifying glass, the single most useful behaviour of a convex lens in daily life.

(i) f=19 cm; (ii) image at infinity (object at focus); (iii) virtual, erect, enlarged image on the same side, shown in the ray diagram.