Light – Reflection and Refraction

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CLASS X Science ~6 marks/year Ch 9 of 13
Light – Reflection and Refraction

Class 10 · Science · NCERT chapter notes · Akanksha Classes

Snapshot
  • Light travels in straight lines; it reflects off mirrors and refracts (bends) when it changes medium.
  • Mirror formula: $\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$ and magnification $m=\dfrac{h'}{h}=-\dfrac{v}{u}$, with $f=\dfrac{R}{2}$.
  • Lens formula: $\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$ and magnification $m=\dfrac{h'}{h}=\dfrac{v}{u}$.
  • Refractive index $n=\dfrac{c}{v}$ measures how much a medium slows and bends light; power of a lens $P=\dfrac{1}{f\,(\text{in m})}$ in dioptres (D).
  • Everything runs on the New Cartesian Sign Convention — get the signs right and the numbers follow.
  • Board weightage: ~6 marks/year — usually one ray-diagram/image-nature question and one numerical using mirror or lens formula (2–3 marks each).
Detailed notes

1. Reflection of light — the two laws

We see objects because light bounces off them into our eyes. A highly polished surface (a mirror) reflects most of the light falling on it. Reflection follows two simple laws:

  • The angle of incidence equals the angle of reflection ($\angle i=\angle r$), both measured from the normal (the line perpendicular to the surface at the point of incidence).
  • The incident ray, the normal, and the reflected ray all lie in the same plane.

These laws hold for every reflecting surface — flat or curved. A plane mirror always gives an image that is virtual, erect, the same size as the object, as far behind the mirror as the object is in front, and laterally inverted (left–right swapped). This chapter explores what changes when the mirror is curved.

2. Spherical mirrors — the vocabulary

A spherical mirror is a small slice of a hollow sphere with one side silvered. Two kinds:

  • Concave mirror — reflecting surface curves inwards (towards the centre of the sphere). It is a converging mirror.
  • Convex mirror — reflecting surface curves outwards. It is a diverging mirror.

Key terms (learn these — questions hide here):

  • Pole (P): the centre of the mirror's reflecting surface — our origin.
  • Centre of curvature (C): the centre of the sphere the mirror is part of. It is not on the mirror. In front for concave, behind for convex.
  • Radius of curvature (R): distance PC, the radius of that sphere.
  • Principal axis: the straight line through P and C (normal to the mirror at P).
  • Principal focus (F): rays parallel to the principal axis, after reflection, meet at F (concave) or appear to come from F (convex).
  • Focal length (f): distance PF.
  • Aperture (MN): the diameter of the reflecting surface. We only study mirrors of small aperture.
$$R=2f\qquad\Longleftrightarrow\qquad f=\dfrac{R}{2}$$

So the focus lies midway between the pole and the centre of curvature. This little relation appears in almost every numerical.

3. Images formed by a concave mirror

What the concave mirror does depends entirely on where you put the object relative to P, F and C. Memorise this table — it is examined directly.

  • Object at infinity → image at F; highly diminished, point-sized; real & inverted.
  • Object beyond C → image between F and C; diminished; real & inverted.
  • Object at C → image at C; same size; real & inverted.
  • Object between C and F → image beyond C; enlarged; real & inverted.
  • Object at F → image at infinity; highly enlarged; real & inverted (image "not formed" on a screen — rays parallel).
  • Object between P and F → image behind the mirror; enlarged; virtual & erect.

Pattern to remember: for a concave mirror the image is real & inverted for every position except when the object is between P and F — only then is it virtual, erect and magnified (this is the "shaving/make-up mirror" case).

Uses of concave mirrors: torch / search-light / vehicle headlight reflectors (source at F gives a parallel beam), shaving mirrors and dentists' mirrors (enlarged erect image), and solar furnaces (concentrate sunlight to produce heat).

4. Images formed by a convex mirror

A convex mirror is far simpler — it has only two cases, and the answer is always the same kind of image:

  • Object at infinity → image at F (behind mirror); highly diminished, point-sized; virtual & erect.
  • Object anywhere between infinity and the pole → image between P and F (behind mirror); diminished; virtual & erect.

So a convex mirror always gives a virtual, erect, diminished image, wherever the object sits.

Uses of convex mirrors: rear-view (wing) mirrors in vehicles — they always give an erect (though smaller) image and have a wider field of view than a plane mirror, so the driver sees more traffic behind. Also used as security mirrors and to view full images of tall buildings in a small mirror.

5. Drawing ray diagrams — the four rules

To locate an image, draw any two of these standard rays from the top of the object; where the reflected rays meet (or appear to meet) is the top of the image. Describe them in words for the board:

  • Rule 1 — parallel ray: a ray parallel to the principal axis, after reflection, passes through F (concave) or appears to diverge from F (convex).
  • Rule 2 — focal ray: a ray passing through F (concave) or directed towards F (convex), after reflection, emerges parallel to the principal axis. (Rule 1 run backwards.)
  • Rule 3 — central-of-curvature ray: a ray through C (or directed towards C) hits the mirror along the normal and retraces its own path.
  • Rule 4 — pole ray: a ray hitting the pole P is reflected making equal angles with the principal axis (obeying $\angle i=\angle r$ about the axis).

If the reflected rays actually cross, the image is real and inverted; if they only appear to meet when produced backwards, the image is virtual and erect.

6. New Cartesian Sign Convention (mirrors)

Every numerical depends on getting signs right. Take the pole P as the origin and the principal axis as the x-axis:

  • The object is always on the left — light travels left → right onto the mirror.
  • All distances are measured from the pole.
  • Distances measured against the incident light (to the left, towards the object) are negative; those along the incident light are positive. So the object distance $u$ is always negative.
  • Heights above the principal axis are positive; heights below are negative.

Consequences worth memorising: for a concave mirror $f$ is negative (focus in front); for a convex mirror $f$ is positive (focus behind). A real image forms in front ($v$ negative); a virtual image forms behind ($v$ positive).

7. Mirror formula and magnification

With object distance $u$, image distance $v$ and focal length $f$ (all signed):

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

Magnification $m$ is how big the image is compared with the object:

$$m=\dfrac{h'}{h}=-\dfrac{v}{u}$$

where $h$ is object height (taken +) and $h'$ image height. Reading the signs: a negative $m$ means a real, inverted image; a positive $m$ means a virtual, erect image. $|m|>1$ is enlarged, $|m|<1$ is diminished. (Always substitute signed values, then let the algebra tell you the result.)

8. NCERT worked Examples — mirrors

NCERT Example 9.1 — convex rear-view mirror

$R=+3.00$ m, $u=-5.00$ m. Convex focus is behind, so $f=+\dfrac{R}{2}=+1.50$ m.

$\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}=\dfrac{1}{1.50}-\dfrac{1}{-5.00}=\dfrac{1}{1.50}+\dfrac{1}{5.00}=\dfrac{5.00+1.50}{7.50}=\dfrac{6.50}{7.50}.$

So $v=\dfrac{+7.50}{6.50}=+1.15$ m — the image is $1.15$ m behind the mirror.

$m=-\dfrac{v}{u}=-\dfrac{1.15}{-5.00}=+0.23.$ The image is virtual, erect and diminished (about $0.23\times$).

NCERT Example 9.2 — concave mirror, screen image

$h=+4.0$ cm, $u=-25.0$ cm, concave so $f=-15.0$ cm.

$\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}=\dfrac{1}{-15.0}-\dfrac{1}{-25.0}=-\dfrac{1}{15.0}+\dfrac{1}{25.0}=\dfrac{-5.0+3.0}{75.0}=\dfrac{-2.0}{75.0}.$

So $v=-37.5$ cm — the screen goes $37.5$ cm in front of the mirror; the image is real.

$h'=-\dfrac{v\,h}{u}=-\dfrac{(-37.5)(+4.0)}{(-25.0)}=-6.0$ cm. The image is inverted and enlarged ($1.5\times$).

9. Refraction of light — bending at a boundary

When light passes obliquely from one transparent medium into another, its direction changes at the boundary. This bending is refraction, and it happens because the speed of light changes between media. (A ray hitting the surface normally, i.e. straight on, does not bend — only its speed changes.)

Everyday effects: a pencil looks broken/displaced in water, a coin in a bowl seems to rise when water is poured, pond bottoms look shallower, letters look raised under a glass slab. Two laws of refraction:

  • The incident ray, the refracted ray and the normal all lie in the same plane.
  • Snell's law: for a given pair of media and a given colour, the ratio $\dfrac{\sin i}{\sin r}$ is a constant.
$$\dfrac{\sin i}{\sin r}=\text{constant}=n_{21}$$

Rule of thumb: going from a rarer to a denser medium (e.g. air → glass), light slows and bends towards the normal. Going from denser to rarer (glass → air), it speeds up and bends away from the normal.

Glass slab: a ray entering a rectangular glass slab bends towards the normal at the first face and away at the second; the emergent ray is parallel to the incident ray, just shifted sideways (lateral displacement), because the two faces are parallel and the bending is equal and opposite.

10. Refractive index

The constant in Snell's law is the refractive index of medium 2 with respect to medium 1, and it equals the ratio of the speeds of light in the two media:

$$n_{21}=\dfrac{\text{speed of light in medium 1}}{\text{speed of light in medium 2}}=\dfrac{v_1}{v_2},\qquad n_{12}=\dfrac{v_2}{v_1}=\dfrac{1}{n_{21}}$$

When medium 1 is vacuum (or air), we get the absolute refractive index $n_m$ of the medium. With $c=3\times10^{8}$ m s$^{-1}$ the speed in vacuum and $v$ the speed in the medium:

$$n_m=\dfrac{c}{v}=\dfrac{\text{speed of light in air/vacuum}}{\text{speed of light in the medium}}$$

Examples: $n_{\text{water}}=1.33$, $n_{\text{crown glass}}=1.52$, $n_{\text{diamond}}=2.42$ (highest in the NCERT table — light slows the most, bends the most). A higher refractive index means optically denser; light travels slower there. Note: optically denser is not the same as mass denser — kerosene ($n=1.44$) is optically denser than water ($n=1.33$) though it is lighter.

11. Spherical lenses — basics

A lens is transparent material bounded by two surfaces, at least one curved.

  • Convex lens (double convex): thicker in the middle, converges light. Also called a converging lens.
  • Concave lens (double concave): thicker at the edges, diverges light. Also called a diverging lens.

Terms: each surface is part of a sphere with a centre of curvature ($C_1$, $C_2$); the line through them is the principal axis; the central point is the optical centre (O) — a ray through O goes straight, undeviated. Each lens has two principal foci $F_1$ and $F_2$ (one on each side, equidistant from O); $f$ = focal length = distance from O to F.

12. Images formed by lenses

Convex lens — depends on object position (note $2F_1$ is the same role as "C" was for mirrors):

  • At infinity → image at $F_2$; highly diminished, point-sized; real & inverted.
  • Beyond $2F_1$ → image between $F_2$ and $2F_2$; diminished; real & inverted.
  • At $2F_1$ → image at $2F_2$; same size; real & inverted.
  • Between $F_1$ and $2F_1$ → image beyond $2F_2$; enlarged; real & inverted.
  • At $F_1$ → image at infinity; highly enlarged; real & inverted (not formed on a screen).
  • Between $F_1$ and O → image on the same side as the object; enlarged; virtual & erect (magnifying-glass case).

Concave lens — like the convex mirror, it is simple and always the same kind of image: for any object position the image is virtual, erect and diminished, formed between $F_1$ and O on the same side as the object.

Lens ray rules: (i) a ray parallel to the axis passes through $F_2$ (convex) / appears to come from $F_1$ (concave); (ii) a ray through $F_1$ (convex) / towards $F_2$ (concave) emerges parallel to the axis; (iii) a ray through the optical centre O goes straight, undeviated.

13. Sign convention, lens formula and magnification

Same New Cartesian convention, but measured from the optical centre O. Result to memorise: convex lens $f$ is positive, concave lens $f$ is negative. The lens formula:

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

(Note the minus sign — different from the mirror formula's plus.) Magnification for a lens:

$$m=\dfrac{h'}{h}=\dfrac{v}{u}$$

(Note: for lenses $m=+\dfrac{v}{u}$ — no leading minus, unlike mirrors.) Again, $m>0$ → virtual & erect; $m<0$ → real & inverted.

14. NCERT worked Examples — lenses

NCERT Example 9.3 — concave lens

Concave lens: $f=-15$ cm. Image (always virtual for a concave lens) at $v=-10$ cm. Find $u$ and $m$.

$\dfrac{1}{u}=\dfrac{1}{v}-\dfrac{1}{f}=\dfrac{1}{-10}-\dfrac{1}{-15}=-\dfrac{1}{10}+\dfrac{1}{15}=\dfrac{-3+2}{30}=\dfrac{1}{-30}.$

So $u=-30$ cm — the object is $30$ cm in front of the lens.

$m=\dfrac{v}{u}=\dfrac{-10}{-30}=+\dfrac{1}{3}=+0.33.$ Positive → virtual, erect, one-third the object's size.

NCERT Example 9.4 — convex lens

$h=+2.0$ cm, convex so $f=+10$ cm, $u=-15$ cm. Find $v$, $h'$, $m$.

$\dfrac{1}{v}=\dfrac{1}{u}+\dfrac{1}{f}=\dfrac{1}{-15}+\dfrac{1}{10}=\dfrac{-2+3}{30}=\dfrac{1}{30}.$ So $v=+30$ cm — image is $30$ cm on the other side, hence real & inverted.

$m=\dfrac{v}{u}=\dfrac{+30}{-15}=-2.$ $h'=m\,h=(-2)(2.0)=-4.0$ cm. The image is real, inverted, enlarged ($2\times$), $4$ cm tall.

15. Power of a lens

A lens of short focal length bends light more strongly. This bending ability is the power:

$$P=\dfrac{1}{f\ (\text{in metres})},\qquad \text{unit}= \text{dioptre (D)},\quad 1\,\text{D}=1\,\text{m}^{-1}$$

1 dioptre is the power of a lens of focal length 1 metre. A convex lens has positive power, a concave lens negative power. For lenses placed in contact, powers add: $P=P_1+P_2+P_3+\dots$ — which is why opticians work in powers, not focal lengths. Example: $P=+2.0$ D means a convex lens of $f=+0.50$ m; $P=-2.5$ D means a concave lens of $f=-0.40$ m.

16. In-text "Questions" — answered

Page 142, Q1. Principal focus of a concave mirror: the point on the principal axis where rays initially parallel to the axis actually meet after reflection.

Q2. $R=20$ cm $\Rightarrow f=\dfrac{R}{2}=10$ cm.

Q3. A concave mirror can give an erect, enlarged image (object between P and F).

Q4. Convex rear-view mirrors give an erect image and a wider field of view, so the driver sees more traffic behind.

Page 145, Q1. $R=32$ cm $\Rightarrow f=\dfrac{R}{2}=16$ cm (convex, so $+16$ cm).

Q2. Concave mirror, real magnified image, $m=-3$ (real ⇒ negative), $u=-10$ cm. $m=-\dfrac{v}{u}\Rightarrow -3=-\dfrac{v}{-10}\Rightarrow v=-30$ cm. The image is 30 cm in front of the mirror (real, inverted).

Page 150, Q1. Light goes air → water (rarer → denser), so it bends towards the normal because it slows down in the denser water.

Q2. $n=1.50$, $c=3\times10^{8}$ m s$^{-1}$. $v=\dfrac{c}{n}=\dfrac{3\times10^{8}}{1.50}=2\times10^{8}$ m s$^{-1}$.

Q3. From the table, diamond ($n=2.42$) has the highest optical density; air ($n=1.0003$) the lowest.

Q4. Light travels fastest in the medium with the lowest $n$: kerosene 1.44, turpentine 1.47, water 1.33 ⇒ fastest in water.

Q5. $n_{\text{diamond}}=2.42$ means light travels $2.42$ times faster in vacuum than in diamond (i.e. $v=\dfrac{c}{2.42}$).

Page 158, Q1. 1 dioptre is the power of a lens whose focal length is 1 metre.

Q2. Convex lens, real inverted image of equal size ⇒ object at $2F$, image at $2F$. Image at $v=+50$ cm, $m=-1\Rightarrow u=-50$ cm. From $\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$: $\dfrac{1}{50}-\dfrac{1}{-50}=\dfrac{2}{50}=\dfrac{1}{25}$, so $f=+25$ cm $=0.25$ m. Needle is 25 cm in front; $P=\dfrac{1}{0.25}=+4$ D.

Q3. Concave lens $f=-2$ m. $P=\dfrac{1}{f}=\dfrac{1}{-2}=-0.5$ D.

17. Exercises — MCQs (Q1–6) answered

  • Q1. A lens cannot be made of (d) Clay — it must be transparent.
  • Q2. Virtual, erect, larger image in a concave mirror ⇒ object (d) between the pole and the principal focus.
  • Q3. Real image of the same size from a convex lens ⇒ object (b) at twice the focal length ($2F$).
  • Q4. $f=-15$ cm for both ⇒ negative $f$ ⇒ (a) both concave (concave mirror and concave lens have negative $f$).
  • Q5. Image always erect, any distance ⇒ (d) either plane or convex.
  • Q6. Reading small letters needs a magnifier ⇒ (c) a convex lens of focal length 5 cm (shorter $f$ → greater magnification).

18. Exercises — descriptive (Q7–9, 13) answered

Q7. Erect image from a concave mirror of $f=15$ cm needs the object between the pole and F, i.e. distance 0 to 15 cm from the mirror. The image is virtual, erect and larger than the object. (Ray diagram: parallel ray reflects through F, ray through pole reflects at equal angle; the two reflected rays diverge and appear to meet behind the mirror.)

Q8. Mirror types with reasons: (a) concave for a car headlight — bulb at F gives a powerful parallel beam; (b) convex for the side/rear-view mirror — erect image, wide field of view; (c) concave for a solar furnace — converges sunlight to a hot focal point.

Q9. Half the convex lens covered with paper still forms a complete image — every part of the lens forms a full image; covering half only reduces the brightness (fewer rays), it does not cut off part of the image.

Q13. A plane mirror's magnification $m=+1$ means the image is the same size as the object ($|m|=1$) and virtual & erect (the $+$ sign).

19. Exercises — numericals (Q10–12, 14–17) solved

Q10. Converging lens $f=+10$ cm, object $5$ cm tall at $u=-25$ cm. $\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}=\dfrac{1}{10}+\dfrac{1}{-25}=\dfrac{5-2}{50}=\dfrac{3}{50}$, so $v=+16.7$ cm. $m=\dfrac{v}{u}=\dfrac{16.7}{-25}=-0.67$; $h'=m\,h=-3.3$ cm. Image is real, inverted, diminished, about $16.7$ cm beyond the lens. (Ray diagram: parallel + central rays.)

Q11. Concave lens $f=-15$ cm, image at $v=-10$ cm (virtual). $\dfrac{1}{u}=\dfrac{1}{v}-\dfrac{1}{f}=\dfrac{1}{-10}-\dfrac{1}{-15}=\dfrac{-3+2}{30}=-\dfrac{1}{30}$, so $u=-30$ cm — object is 30 cm in front of the lens.

Q12. Convex mirror $f=+15$ cm, $u=-10$ cm. $\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}=\dfrac{1}{15}+\dfrac{1}{10}=\dfrac{2+3}{30}=\dfrac{5}{30}=\dfrac{1}{6}$, so $v=+6$ cm. $m=-\dfrac{v}{u}=-\dfrac{6}{-10}=+0.6$. Image is 6 cm behind the mirror: virtual, erect, diminished.

Q14. Convex mirror, $R=30$ cm ⇒ $f=+15$ cm, object $5.0$ cm at $u=-20$ cm. $\dfrac{1}{v}=\dfrac{1}{15}+\dfrac{1}{20}=\dfrac{4+3}{60}=\dfrac{7}{60}$, so $v=+8.6$ cm. $m=-\dfrac{v}{u}=-\dfrac{8.6}{-20}=+0.43$; $h'=0.43\times5.0=+2.2$ cm. Image is virtual, erect, diminished ($\approx2.2$ cm) about $8.6$ cm behind the mirror.

Q15. Concave mirror $f=-18$ cm, object $7.0$ cm at $u=-27$ cm. $\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}=\dfrac{1}{-18}-\dfrac{1}{-27}=\dfrac{-3+2}{54}=-\dfrac{1}{54}$, so $v=-54$ cm — screen at 54 cm in front. $m=-\dfrac{v}{u}=-\dfrac{-54}{-27}=-2$; $h'=-2\times7.0=-14$ cm. Image is real, inverted, enlarged ($14$ cm).

Q16. $P=-2.0$ D ⇒ $f=\dfrac{1}{P}=\dfrac{1}{-2.0}=-0.5$ m $=-50$ cm. Negative ⇒ concave (diverging) lens.

Q17. $P=+1.5$ D ⇒ $f=\dfrac{1}{1.5}=+0.67$ m $=+67$ cm (approx). Positive ⇒ convex / converging lens.

20. Common mistakes to avoid

  • Forgetting the sign convention: object distance $u$ is always negative; concave $f$ negative, convex $f$ positive.
  • Mixing the two formulae: mirror is $\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$ (plus), lens is $\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$ (minus).
  • Mixing magnification: mirror $m=-\dfrac{v}{u}$ (minus), lens $m=+\dfrac{v}{u}$ (plus).
  • Using $f=R$ instead of $f=\dfrac{R}{2}$ for a mirror.
  • Saying a ray bends when it hits a surface normally — it doesn't (only its speed changes).
  • Confusing optical density with mass density.
  • Putting power in dioptres while $f$ is in cm — $f$ must be in metres for $P=\dfrac{1}{f}$.

21. Quick revision checklist

  • Laws of reflection: $\angle i=\angle r$; rays and normal coplanar.
  • $f=\dfrac{R}{2}$; concave converges, convex diverges & always gives virtual erect diminished image.
  • Mirror: $\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$, $m=-\dfrac{v}{u}$. Concave $f<0$, convex $f>0$.
  • Refraction: Snell's law $\dfrac{\sin i}{\sin r}=n_{21}=\dfrac{v_1}{v_2}$; $n=\dfrac{c}{v}$; rarer→denser bends towards normal.
  • Lens: $\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$, $m=\dfrac{v}{u}$. Convex $f>0$, concave $f<0$; concave lens always virtual erect diminished.
  • Power $P=\dfrac{1}{f(\text{m})}$ in dioptres; powers in contact add.
  • $m<0$ real & inverted; $m>0$ virtual & erect.
Practice MCQs
1. The focal length of a spherical mirror of radius of curvature $24$ cm is:
  1. $48$ cm
  2. $24$ cm
  3. $12$ cm
  4. $6$ cm
Answer: (C) $f=\dfrac{R}{2}=\dfrac{24}{2}=12$ cm.
2. A convex mirror always forms an image that is:
  1. real and inverted
  2. virtual, erect and diminished
  3. virtual and enlarged
  4. real and enlarged
Answer: (B) for every object position a convex mirror gives a virtual, erect, diminished image.
3. For a concave mirror, a real and same-sized image is formed when the object is:
  1. at F
  2. between P and F
  3. at C
  4. at infinity
Answer: (C) object at C ⇒ image at C, same size, real & inverted.
4. Light bends towards the normal when it passes from:
  1. denser to rarer medium
  2. rarer to denser medium
  3. glass to air
  4. water to air
Answer: (B) entering a denser medium light slows and bends towards the normal.
5. The refractive index of glass is $1.5$ and $c=3\times10^{8}$ m s$^{-1}$. Speed of light in glass is:
  1. $4.5\times10^{8}$
  2. $3\times10^{8}$
  3. $2\times10^{8}$
  4. $1.5\times10^{8}$
Answer: (C) $v=\dfrac{c}{n}=\dfrac{3\times10^{8}}{1.5}=2\times10^{8}$ m s$^{-1}$.
6. The SI unit of power of a lens is:
  1. metre
  2. dioptre
  3. watt
  4. candela
Answer: (B) dioptre (D), where $1$ D $=1$ m$^{-1}$.
7. A lens of focal length $-50$ cm is a:
  1. convex lens of power $+2$ D
  2. concave lens of power $-2$ D
  3. convex lens of power $-2$ D
  4. concave lens of power $+0.5$ D
Answer: (B) $f=-0.5$ m ⇒ concave, $P=\dfrac{1}{-0.5}=-2$ D.
8. For a convex lens, a magnifying-glass image (virtual, erect, enlarged) forms when the object is:
  1. beyond $2F_1$
  2. at $2F_1$
  3. between $F_1$ and $2F_1$
  4. between $F_1$ and O
Answer: (D) object between the focus and the optical centre.
9. A negative value of magnification for a mirror indicates the image is:
  1. virtual and erect
  2. real and inverted
  3. diminished
  4. at infinity
Answer: (B) $m<0$ ⇒ real and inverted.
10. The mirror used in a car headlight reflector is:
  1. plane
  2. convex
  3. concave
  4. cylindrical
Answer: (C) concave — a source at the focus gives a strong parallel beam.
11. From the NCERT table, the medium with the highest optical density (largest $n$) is:
  1. water
  2. crown glass
  3. diamond
  4. kerosene
Answer: (C) diamond, $n=2.42$.
12. A ray of light hitting a glass slab normally (perpendicularly) will:
  1. bend towards the normal
  2. bend away from the normal
  3. not bend, but change speed
  4. be totally reflected
Answer: (C) at normal incidence there is no bending; only the speed changes.
Assertion–Reason
A: A convex mirror is used as a rear-view mirror in vehicles.   R: A convex mirror always forms a virtual, erect, diminished image and has a wide field of view.
Answer: Both A and R are true, and R correctly explains A — the erect, diminished, wide-angle image is exactly why convex mirrors suit rear-view use.
A: The focal length of a concave lens is taken as negative.   R: A concave lens forms a real, inverted image for distant objects.
Answer: A is true, R is false — a concave lens always forms a virtual, erect, diminished image, never a real one.
Previous-year questions
Q1. An object $5.0$ cm tall is placed $20$ cm in front of a convex mirror of radius of curvature $30$ cm. Find the position, nature and size of the image. (CBSE, 3 marks)
Answer: $f=+15$ cm, $u=-20$ cm. $\dfrac{1}{v}=\dfrac{1}{15}+\dfrac{1}{20}=\dfrac{7}{60}\Rightarrow v=+8.6$ cm. $m=-\dfrac{v}{u}=+0.43$, $h'=+2.2$ cm. Image: virtual, erect, diminished, $8.6$ cm behind the mirror.
Q2. State the laws of refraction of light. Define refractive index of a medium. (CBSE, 3 marks)
Answer: (i) Incident ray, refracted ray and normal lie in one plane. (ii) Snell's law: $\dfrac{\sin i}{\sin r}=$ constant for a given pair of media. Refractive index $n=\dfrac{c}{v}=\dfrac{\text{speed in vacuum}}{\text{speed in medium}}$ — how much the medium slows light.
Q3. A concave mirror produces a three-times magnified real image of an object placed $10$ cm in front of it. Where is the image located? (CBSE, 2 marks)
Answer: Real ⇒ $m=-3$. $m=-\dfrac{v}{u}\Rightarrow -3=-\dfrac{v}{-10}\Rightarrow v=-30$ cm — image is $30$ cm in front of the mirror (real, inverted).
Q4. A doctor prescribes a corrective lens of power $+1.5$ D. Find its focal length and state whether it is converging or diverging. (CBSE, 2 marks)
Answer: $f=\dfrac{1}{P}=\dfrac{1}{1.5}=+0.67$ m ($\approx+67$ cm). Positive power ⇒ convex (converging) lens.
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