NCERT Exemplar Solutions Class 10 Science Light Reflection and Refraction — Short and Long Answer Questions Q20–38

NCERT Exemplar Problems, Class 10 Science — Light: Reflection and Refraction, Short Answer Questions 20 to 29 and Long Answer Questions 30 to 38. All 19 questions solved.

Nine of these nineteen questions ask for a ray diagram, and five more are numerical. Both kinds go wrong for the same reason — a mishandled sign — so it is worth fixing the convention before anything else.

Distances are measured from the pole (or optical centre), and anything measured against the direction of the incident light is negative. In practice, for light arriving from the left:

Quantity	Sign
Object distance $u$	always negative
Real image, mirror ($v$ in front)	negative
Real image, lens ($v$ on the far side)	positive
Concave mirror $f$ / convex lens $f$	negative / positive
Convex mirror $f$ / concave lens $f$	positive / negative
Height above the axis $h$	positive; below, negative

And the two working equations, which differ by one sign:

$$\text{mirror: } \frac{1}{f} = \frac{1}{v} + \frac{1}{u}, \quad m = -\frac{v}{u} \qquad\qquad \text{lens: } \frac{1}{f} = \frac{1}{v} – \frac{1}{u}, \quad m = \frac{v}{u}$$

Key insight. Every ray diagram in this band is built from the same two or three standard rays, and you never need more than two. For a mirror: a ray parallel to the axis reflects through F, and a ray through C returns along itself. For a lens: a ray parallel to the axis passes through F on the far side, and a ray through the optical centre goes straight on. Where the two emergent rays meet is the image — and if they only meet when produced backwards, the image is virtual.

Short Answer Questions

Question 20

Identify the device used as a spherical mirror or lens in following cases, when the image formed is virtual and erect in each case.

(a) Object is placed between device and its focus, image formed is enlarged and behind it. (b) Object is placed between the focus and device, image formed is enlarged and on the same side as that of the object. (c) Object is placed between infinity and device, image formed is diminished and between focus and optical centre on the same side as that of the object. (d) Object is placed between infinity and device, image formed is diminished and between pole and focus, behind it.

Solution. Two clues separate the four devices, and taking them in order makes the question mechanical.

First: which side is the image on? The words behind it mean the image is on the far side of the surface from the object — and for a virtual image that only happens with a mirror. On the same side as the object means a lens. So (a) and (d) are mirrors, (b) and (c) are lenses.

Second: is the image enlarged or diminished? Only the two converging devices — concave mirror and convex lens — can enlarge. The diverging pair always diminish.

Putting the two together:

(a) Mirror, enlarged → concave mirror. An object nearer than F gives a magnified virtual image behind it. This is the shaving mirror.

(b) Lens, enlarged → convex lens. An object nearer than F gives a magnified virtual image on the same side. This is the magnifying glass.

(c) Lens, diminished, image always between F and the optical centre → concave lens. It gives this same result for every object position, which is what Question 33 shows.

(d) Mirror, diminished, image always between P and F behind → convex mirror. Again true for every object position, as Question 34 shows.

Notice the symmetry: (a) and (b) are the two magnifiers, each used with the object inside the focus; (c) and (d) are the two diminishers, which behave the same way whatever you do.

Question 21

Why does a light ray incident on a rectangular glass slab immersed in any medium emerges parallel to itself? Explain using a diagram.

Solution. Because the two refracting faces of a rectangular slab are parallel, so the bending at the second surface exactly undoes the bending at the first.

At the first surface the ray passes from the surrounding medium into the denser glass, so it bends towards the normal:

$$n = \frac{\sin i}{\sin r}$$

At the second surface it passes from glass back into the same surrounding medium. The two faces are parallel, so their normals are parallel, and the angle at which the ray meets the second surface from the inside is the same $r$ as the angle of refraction at the first. Applying Snell’s law in the reverse direction,

$$\frac{\sin r}{\sin i’} = \frac{1}{n} \quad \Rightarrow \quad \sin i’ = n \sin r = \sin i \quad \Rightarrow \quad i’ = i$$

The emergent ray therefore makes the same angle with the normal as the incident ray did, on the same side — so it is parallel to the incident ray.

The one thing that is not restored is the ray’s position: the detour through the glass shifts it sideways. That offset is the lateral displacement, and it grows with the thickness of the slab and with the angle of incidence.

The phrase immersed in any medium matters. The argument never used the value of $n$ — only that the medium is the same on both sides. So the result holds in air, in water, in oil; only the size of the shift changes.

Question 22

A pencil when dipped in water in a glass tumbler appears to be bent at the interface of air and water. Will the pencil appear to be bent to the same extent, if instead of water we use liquids like, kerosene or turpentine. Support your answer with reason.

Solution. No. The apparent bending will be greater in kerosene and greater still in turpentine.

The pencil looks bent because light coming from the submerged part is refracted as it leaves the liquid and enters the air, so the eye sees that part displaced from where it really is. How far it is displaced depends on how much the light is bent — and that depends on the refractive index of the liquid:

$$n = \frac{\sin i}{\sin r}$$

A larger $n$ bends the light more, and so shifts the apparent position more.

Liquid	Refractive index
Water	1.33
Kerosene	1.44
Turpentine	1.47

Both kerosene and turpentine have a higher refractive index than water, so both bend the emerging light more than water does and the pencil appears more sharply bent in them.

The general statement, which is also the answer to Question 15, is that the extent of bending is fixed by the refractive index of the medium — equivalently by how much light is slowed in it, since $n = c/v$.

Question 23

How is the refractive index of a medium related to the speed of light? Obtain an expression for refractive index of a medium with respect to another in terms of speed of light in these two media?

Solution. Light travels fastest in vacuum, at $c = 3 \times 10^8\ \mathrm{m\,s^{-1}}$, and more slowly in any material. The absolute refractive index of a medium measures exactly that slowing:

$$n = \frac{c}{v} = \frac{\text{speed of light in vacuum}}{\text{speed of light in the medium}}$$

Since $v < c$ always, $n$ is always greater than 1, and a larger $n$ means a slower speed — which is why a medium with a high refractive index is called optically denser.

Now the relative refractive index. Take two media, 1 and 2, in which light travels at $v_1$ and $v_2$. Their absolute indices are

$$n_1 = \frac{c}{v_1}, \qquad n_2 = \frac{c}{v_2}$$

The refractive index of medium 2 with respect to medium 1 is the ratio of the two:

$${}_1n_2 = \frac{n_2}{n_1} = \frac{c/v_2}{c/v_1} = \frac{v_1}{v_2}$$

so

$$\boxed{\ {}_1n_2 = \frac{v_1}{v_2}\ }$$

Note the inversion that happens in that last step: the index of 2 with respect to 1 carries $v_1$ on top, the speed in the first medium. It is the single most common slip in this derivation.

The result also shows that ${}_1n_2 \times {}_2n_1 = 1$ — the two relative indices are reciprocals, which is the fact Question 24 relies on.

Question 24

Refractive index of diamond with respect to glass is 1.6 and absolute refractive index of glass is 1.5. Find out the absolute refractive index of diamond.

Solution. “With respect to glass” means a relative refractive index, and by the definition from Question 23 it is the ratio of the two absolute indices:

$${}_g n_d = \frac{n_d}{n_g}$$

Rearranging for what is wanted:

$$n_d = {}_g n_d \times n_g = 1.6 \times 1.5 = \mathbf{2.40}$$

So the absolute refractive index of diamond is $2.40$ — which is close to its true value of about 2.42, the highest of any common material and the reason diamonds sparkle.

It is worth checking the answer against the speeds, which makes the size of the number concrete:

$$v_{\text{diamond}} = \frac{c}{n_d} = \frac{3 \times 10^8}{2.40} = 1.25 \times 10^8\ \mathrm{m\,s^{-1}}$$

Light travels in diamond at less than half its speed in vacuum.

The common error here is to divide rather than multiply — to write $1.6 / 1.5$. A quick sanity check catches it: diamond is far more refractive than glass, so its absolute index must be larger than 1.5, and $1.6/1.5 \approx 1.07$ is smaller.

Question 25

A convex lens of focal length 20 cm can produce a magnified virtual as well as real image. Is this a correct statement? If yes, where shall the object be placed in each case for obtaining these images?

Solution. Yes, the statement is correct. A convex lens is the one device that magnifies in two different ways, and the object position decides which.

For a magnified virtual image, place the object between the optical centre and the focus — that is, within 20 cm of the lens. The emergent rays still diverge, so they never meet; produced backwards they appear to come from an enlarged, erect image on the same side as the object. This is the magnifying glass, and it is case (a) of Question 31.

For a magnified real image, place the object between F and 2F — that is, between 20 cm and 40 cm from the lens. Now the emergent rays actually converge, and they meet beyond 2F to give an enlarged, inverted image that can be caught on a screen. This is case (b) of Question 31, and it is how a projector works.

The boundaries are worth understanding rather than memorising. At exactly 20 cm the object is at F and the emergent rays are parallel — the image is at infinity, the changeover point between the two cases. At exactly 40 cm the object is at 2F and the image is the same size, so beyond 40 cm it is diminished.

$$\underbrace{0 \to 20\ \mathrm{cm}}_{\text{virtual, magnified}} \quad\Big|\quad \underbrace{20 \to 40\ \mathrm{cm}}_{\text{real, magnified}} \quad\Big|\quad \underbrace{> 40\ \mathrm{cm}}_{\text{real, diminished}}$$

Question 26

Sudha finds out that the sharp image of the window pane of her science laboratory is formed at a distance of 15 cm from the lens. She now tries to focus the building visible to her outside the window instead of the window pane without disturbing the lens. In which direction will she move the screen to obtain a sharp image of the building? What is the approximate focal length of this lens?

Solution. The governing fact is that for a converging lens, as the object moves further away, its image moves closer to the lens — approaching the focus as the object approaches infinity.

The building is much further away than the window pane. So its image forms nearer to the lens than 15 cm, and Sudha must move the screen towards the lens.

For the focal length: the window pane is already fairly distant compared with the lens, so its image at 15 cm is already close to the limiting position. A building further still is effectively at infinity, and the image of an object at infinity forms at the focus. So

$$f \approx 15\ \mathrm{cm}$$

— strictly a little less than 15 cm, since the window pane is not truly at infinity, but 15 cm is the approximation the question asks for.

The lens formula makes the direction of movement exact. With $f = 15$:

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

and as $u \to -\infty$ the term $1/u \to 0$, so $v \to f = 15$ from above. Every increase in object distance decreases $v$.

This is also the standard laboratory method for measuring focal length, and it is the method behind Questions 10 and 19: focus a distant object and measure the image distance.

Question 27

How are power and focal length of a lens related? You are provided with two lenses of focal length 20 cm and 40 cm respectively. Which lens will you use to obtain more convergent light?

Solution. The power of a lens is the reciprocal of its focal length, with the focal length in metres:

$$P = \frac{1}{f}, \qquad P \propto \frac{1}{f}$$

and the unit of power is the dioptre (D), the power of a lens of focal length one metre.

Because the relation is inverse, a shorter focal length means a greater power — and greater power means the lens bends light more strongly, that is, converges it more. Computing both:

$$P_1 = \frac{1}{0.20} = +5\ \mathrm{D}, \qquad P_2 = \frac{1}{0.40} = +2.5\ \mathrm{D}$$

The 20 cm lens has twice the power of the 40 cm lens, so the 20 cm lens gives the more convergent light.

The physical picture behind the algebra is worth keeping: a short focal length means parallel rays are brought to a point after a short distance, which requires bending them through a large angle. A “powerful” lens is a strongly curved, fat one; a weak lens is nearly flat.

Note that both focal lengths here are positive, so both lenses are convex and both converge. A concave lens would have negative $f$ and negative power, and would diverge.

Question 28

Under what condition in an arrangement of two plane mirrors, incident ray and reflected ray will always be parallel to each other, whatever may be angle of incidence. Show the same with the help of diagram.

Solution. The condition is that the two plane mirrors are placed at right angles, $90^\circ$, to each other.

The proof. Let the ray strike mirror 1 at an angle of incidence $\theta$ from its normal. By the law of reflection it leaves at $\theta$ as well, so it has been turned through $180^\circ – 2\theta$; equivalently the deviation at the first mirror is $180^\circ – 2\theta$.

That reflected ray now travels to mirror 2. Because the mirrors are perpendicular, their normals are perpendicular too, so the angle this ray makes with the second normal is $90^\circ – \theta$. Reflecting again deviates it by $180^\circ – 2(90^\circ – \theta) = 2\theta$.

The total deviation is therefore

$$(180^\circ – 2\theta) + 2\theta = 180^\circ$$

and the $\theta$ has cancelled. A total deviation of $180^\circ$ means the final ray travels in exactly the opposite direction to the incident ray — that is, parallel to it — and since $\theta$ dropped out, this holds for every angle of incidence.

This arrangement is called a retroreflector, and it is why bicycle reflectors and road studs send light straight back to its source whatever direction it arrives from. (Three mirrors at mutual right angles do the same in three dimensions, which is how the reflectors left on the Moon return laser light to Earth.)

Question 29

Draw a ray diagram showing the path of rays of light when it enters with oblique incidence (i) from air into water; (ii) from water into air.

Solution. The two cases are opposites, and the rule that decides both is the same: light bends towards the normal on entering a denser medium and away from the normal on entering a rarer one.

Water ($n = 1.33$) is denser than air ($n = 1.00$), so:

(i) Air into water. Water is the denser medium, so the ray slows down and bends towards the normal: the angle of refraction is smaller than the angle of incidence, $r < i$.

(ii) Water into air. Air is the rarer medium, so the ray speeds up and bends away from the normal: $r > i$.

The two diagrams are each other’s reverse, which is not a coincidence — it is the reversibility of light paths. Run case (i) backwards and you have case (ii).

One consequence worth noting for case (ii): since $r > i$ always, there is an angle of incidence in water at which $r$ would reach $90^\circ$. Beyond it no refracted ray escapes at all and the light is entirely reflected back into the water. That is total internal reflection, and it has no counterpart in case (i).

Long Answer Questions

Question 30

Draw ray diagrams showing the image formation by a concave mirror when an object is placed (a) between pole and focus of the mirror (b) between focus and centre of curvature of the mirror (c) at centre of curvature of the mirror (d) a little beyond centre of curvature of the mirror (e) at infinity

Solution. Each diagram uses the same two construction rays: one parallel to the principal axis, which reflects through F; and one through C, which strikes the mirror along the normal and returns along itself. Where the two reflected rays meet is the image — and where they meet only when produced behind the mirror, the image is virtual and is drawn dashed.

Reading the five together shows the pattern: as the object moves in from infinity towards the mirror, the image moves out from F, growing all the while — until the object reaches F, where the reflected rays are parallel and no image forms. Cross that point and the image jumps behind the mirror and becomes virtual and erect.

Object position	Image position	Nature
(a) Between P and F	Behind the mirror	Virtual, erect, magnified
(b) Between F and C	Beyond C	Real, inverted, magnified
(c) At C	At C	Real, inverted, same size
(d) Just beyond C	Between F and C	Real, inverted, diminished
(e) At infinity	At F	Real, inverted, point sized

Case (a) is the shaving mirror and case (e) is the method for measuring focal length in Question 10. Cases (b) and (a) between them are why Question 3 of the multiple-choice band has two correct options.

Question 31

Draw ray diagrams showing the image formation by a convex lens when an object is placed (a) between optical centre and focus of the lens (b) between focus and twice the focal length of the lens (c) at twice the focal length of the lens (d) at infinity (e) at the focus of the lens

Solution. The two construction rays for a lens are: one parallel to the principal axis, which passes through F′ on the far side; and one through the optical centre, which goes straight on undeviated.

Object position	Image position	Nature
(a) Between O and F	Same side as the object	Virtual, erect, magnified
(b) Between F and 2F	Beyond 2F′	Real, inverted, magnified
(c) At 2F	At 2F′	Real, inverted, same size
(d) At infinity	At F′	Real, inverted, point sized
(e) At F	At infinity	Real, inverted, highly magnified

Case (e) is the one worth pausing on. With the object exactly at F, the two emergent rays come out parallel to each other and never meet, so no image forms at any finite distance — the image is said to be at infinity. It is the boundary between the real images of (b) to (d) and the virtual image of (a), which is why moving an object slowly towards a magnifying glass makes the image grow without limit and then flip.

Compare this table with the concave mirror’s in Question 30: they are the same sequence of five outcomes, with “behind the mirror” replaced by “same side as the object”. A converging lens and a converging mirror behave alike; only the geometry of where the image lands differs.

Question 32

Write laws of refraction. Explain the same with the help of ray diagram, when a ray of light passes through a rectangular glass slab.

Solution. The two laws of refraction:

First law. The incident ray, the refracted ray and the normal to the interface at the point of incidence all lie in the same plane.

Second law (Snell’s law). For light passing between two given media, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant:

$$\frac{\sin i}{\sin r} = \text{constant} = {}_1n_2$$

That constant is the refractive index of the second medium with respect to the first, and by Question 23 it equals $v_1 / v_2$.

Applied to a glass slab. The diagram is the one drawn in Question 21 above. The ray meets the top face at angle $i$ and is refracted into the glass at the smaller angle $r$, obeying $\sin i / \sin r = n$ — that is the second law at the first surface, and the fact that the bent ray stays in the plane of the page is the first law.

Inside the glass the ray travels in a straight line, because the medium is uniform. At the bottom face it meets the boundary from the inside at the same angle $r$, since the two faces are parallel, and emerges into air at

$$\frac{\sin r}{\sin i’} = \frac{1}{n} \quad\Rightarrow\quad i’ = i$$

so the emergent ray is parallel to the incident ray, displaced laterally.

The slab is a good demonstration precisely because it applies the laws twice in opposite directions, and the cancellation makes the result easy to check with a ruler: lay the incident and emergent rays alongside each other and they are parallel.

Question 33

Draw ray diagrams showing the image formation by a concave lens when an object is placed (a) at the focus of the lens (b) between focus and twice the focal length of the lens (c) beyond twice the focal length of the lens

Solution. A concave lens is diverging, so the emergent rays always spread apart and can never meet in front of the lens. Produced backwards they appear to come from a point on the same side as the object, which is why the image is always virtual.

The striking feature is that all three cases give the same kind of image: virtual, erect and diminished, always located between the optical centre and F on the object’s side. Working the lens formula with $f = -F$ confirms it in general — for an object at distance $d$,

$$\frac{1}{v} = \frac{1}{-F} + \frac{1}{-d} \quad\Rightarrow\quad v = -\frac{Fd}{F+d}, \qquad m = \frac{F}{F+d}$$

Since $F$ and $d$ are both positive, $v$ is always negative (same side, virtual) and $m$ is always between 0 and 1 (erect and diminished), whatever $d$ is.

Object at	Image distance	Magnification
(a) F	$F/2$	$1/2$
(b) Between F and 2F	between $F/2$ and $2F/3$	between $1/2$ and $1/3$
(c) Beyond 2F	approaching $F$	below $1/3$, falling towards 0

So as the object recedes, the image shrinks and creeps towards F, but never passes it. This unchanging behaviour is exactly what makes the concave lens identifiable in Question 20(c), and it is why concave lenses are used to correct short sight, where a fixed diminishing effect is wanted.

Question 34

Draw ray diagrams showing the image formation by a convex mirror when an object is placed (a) at infinity (b) at finite distance from the mirror

Solution. A convex mirror is diverging. The reflected rays always spread apart, so they never meet in front of the mirror; produced backwards they appear to come from a point behind it, and the image is always virtual.

(a) Object at infinity. The incoming rays are parallel. After reflection they diverge as though coming from the focus, so the image is a point at F behind the mirror — virtual, erect and highly diminished.

(b) Object at a finite distance. The image is virtual, erect and diminished, and lies between the pole and the focus behind the mirror.

As with the concave lens, the behaviour never changes. For an object at distance $d$ with focal length $F$,

$$v = \frac{Fd}{F+d}, \qquad m = \frac{F}{F+d}$$

Both are positive, and $m$ is always less than 1 — so the image is always behind the mirror, always erect, always diminished. That is the content of Question 9, and as the object recedes the image shrinks towards the focus without ever reaching it.

The whole image of everything in front of the mirror is therefore squeezed into the small region between P and F, which is exactly why a convex mirror has such a wide field of view and is used for rear view mirrors and on blind corners.

Question 35

The image of a candle flame formed by a lens is obtained on a screen placed on the other side of the lens. If the image is three times the size of the flame and the distance between lens and image is 80 cm, at what distance should the candle be placed from the lens? What is the nature of the image at a distance of 80 cm and the lens?

Solution. The image is caught on a screen, so it is real; and a real image formed by a lens is inverted. Fix the signs from that before doing anything else:

$$v = +80\ \mathrm{cm}, \qquad m = -3$$

The magnification is negative because the image is inverted, and its size is 3 for “three times the size”.

Finding the object distance. For a lens, $m = v/u$, so

$$-3 = \frac{80}{u} \quad\Rightarrow\quad u = -\frac{80}{3} = -26.7\ \mathrm{cm}$$

The candle is 26.7 cm from the lens, on the incoming side.

Finding the focal length, to identify the lens:

$$\frac{1}{f} = \frac{1}{v} – \frac{1}{u} = \frac{1}{80} – \frac{1}{-80/3} = \frac{1}{80} + \frac{3}{80} = \frac{4}{80} = \frac{1}{20}$$

$$f = +20\ \mathrm{cm}$$

A positive focal length means the lens is convex (converging).

Nature of the image: real, inverted and magnified three times.

The answer passes a useful check. With $f = 20$, the object at 26.7 cm lies between F (20 cm) and 2F (40 cm) — and that is precisely the region which Question 31(b) says gives a real, inverted, magnified image. The two results agree.

Question 36

Size of image of an object by a mirror having a focal length of 20 cm is observed to be reduced to 1/3rd of its size. At what distance the object has been placed from the mirror? What is the nature of the image and the mirror?

Solution. The image is one-third the size, so $|m| = 1/3$. The question does not say whether the mirror is concave or convex, so both possibilities have to be tested — and it turns out both work.

Case 1: a concave mirror, $f = -20\ \mathrm{cm}$. A concave mirror gives a diminished image only when that image is real and inverted, so $m = -1/3$:

$$m = -\frac{v}{u} = -\frac{1}{3} \quad\Rightarrow\quad v = \frac{u}{3}$$

$$\frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{3}{u} + \frac{1}{u} = \frac{4}{u} \quad\Rightarrow\quad u = 4f = -80\ \mathrm{cm}$$

So the object is at 80 cm, the image is at $v = -26.7$ cm — real and inverted. This is the key’s answer.

Case 2: a convex mirror, $f = +20\ \mathrm{cm}$. A convex mirror always gives a virtual erect image, so $m = +1/3$:

$$-\frac{v}{u} = \frac{1}{3} \quad\Rightarrow\quad v = -\frac{u}{3}, \qquad \frac{1}{20} = -\frac{3}{u} + \frac{1}{u} = -\frac{2}{u} \quad\Rightarrow\quad u = -40\ \mathrm{cm}$$

So the object is at 40 cm, the image at $v = +13.3$ cm behind the mirror — virtual and erect.

Both are consistent with everything the question states. The question is therefore under-determined: “a focal length of 20 cm” does not say which sign is meant, and an image “reduced to 1/3rd” happens in both cases. The key takes the concave reading, so that is the answer to give — but a complete solution should say why, and the convex case is worth working because it is equally instructive.

Question 37

Define power of a lens. What is its unit? One student uses a lens of focal length 50 cm and another of –50 cm. What is the nature of the lens and its power used by each of them?

Solution. Definition. The power of a lens is its ability to converge or diverge a beam of light, and it is measured as the reciprocal of the focal length in metres:

$$P = \frac{1}{f\ (\text{in metres})}$$

Unit. The dioptre, symbol D. One dioptre is the power of a lens whose focal length is one metre. Since $f$ must be in metres, converting from centimetres is the first step of every calculation.

The first student, $f = 50\ \mathrm{cm} = +0.50\ \mathrm{m}$:

$$P = \frac{1}{0.50} = +2\ \mathrm{D}$$

A positive focal length and positive power mean a convex (converging) lens.

The second student, $f = -50\ \mathrm{cm} = -0.50\ \mathrm{m}$:

$$P = \frac{1}{-0.50} = -2\ \mathrm{D}$$

A negative focal length and negative power mean a concave (diverging) lens.

The sign convention is doing real work here: the two lenses have the same focal length in magnitude and therefore the same strength, but they act in opposite senses. This is exactly how spectacle prescriptions are written — a $+2\ \mathrm{D}$ lens corrects long sight by adding convergence, a $-2\ \mathrm{D}$ lens corrects short sight by adding divergence.

Question 38

A student focussed the image of a candle flame on a white screen using a convex lens. He noted down the position of the candle, screen and the lens as under: position of candle = 12.0 cm, position of convex lens = 50.0 cm, position of the screen = 88.0 cm.

(i) What is the focal length of the convex lens? (ii) Where will the image be formed if he shifts the candle towards the lens at a position of 31.0 cm? (iii) What will be the nature of the image formed if he further shifts the candle towards the lens? (iv) Draw a ray diagram to show the formation of the image in case (iii) as said above.

Solution. The positions are readings on an optical bench, so the distances have to be worked out as differences before any formula is used.

$$u = -(50.0 – 12.0) = -38.0\ \mathrm{cm}, \qquad v = +(88.0 – 50.0) = +38.0\ \mathrm{cm}$$

(i) Substituting in the lens formula:

$$\frac{1}{f} = \frac{1}{v} – \frac{1}{u} = \frac{1}{38} + \frac{1}{38} = \frac{2}{38} \quad\Rightarrow\quad f = \mathbf{19.0\ cm}$$

There is a quicker route worth knowing: when the object and image distances are equal, the object is at $2F$ and the image at $2F’$, so $f = 38/2 = 19$ cm immediately. That is case (c) of Question 31 — image the same size as the object.

(ii) The candle moves to 31.0 cm, so now

$$u = -(50.0 – 31.0) = -19.0\ \mathrm{cm} = -f$$

The object is exactly at the focus. Then

$$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{19} – \frac{1}{19} = 0$$

so $v$ is infinite: the emergent rays are parallel and the image is formed at infinity. No image can be caught on the screen at any distance — this is case (e) of Question 31.

(iii) Shifting the candle still closer puts the object inside the focal length. The emergent rays then diverge, and the image becomes virtual, erect and magnified, on the same side as the candle — the magnifying-glass case, (a) of Question 31.

(iv) The ray diagram is panel (a) of the convex-lens set in Question 31 above: the ray parallel to the axis refracts through F′, the ray through the optical centre goes straight on, the two diverge, and their backward extensions (dashed) meet to locate the enlarged erect virtual image on the object’s side.

Common mistakes

• Forgetting that a screen image is real and inverted, in Questions 35 and 38. “Obtained on a screen” fixes $m$ as negative. Taking $m = +3$ in Question 35 puts the object on the wrong side and gives the wrong focal length.
• Using centimetres in $P = 1/f$, in Question 37. The focal length must be in metres, or the power comes out a hundred times too large.
• Inverting the relative refractive index, in Questions 23 and 24. ${}_1n_2 = v_1/v_2$ — the speed in the first medium on top — and $n_d = {}_g n_d \times n_g$, a multiplication, not a division.
• Reading optical-bench positions as distances, in Question 38. The candle at 12.0 cm and the lens at 50.0 cm give $u = -38$ cm, not $-12$ cm. Always subtract.
• Drawing a virtual image with solid lines. In Questions 30(a), 33 and 34 the rays never actually meet; it is their backward extensions that do, and those must be dashed or the diagram claims something false.
• Assuming a diminished image means a convex mirror, in Question 36. A concave mirror also diminishes, whenever the object is beyond C — which is why that question has two consistent answers.
• Stopping at one case in Question 25. A convex lens magnifies in two separate ranges, giving a virtual image inside F and a real one between F and 2F.
• Quoting Snell’s law without the plane condition, in Question 32. There are two laws of refraction, and the first — that the incident ray, refracted ray and normal are coplanar — is the one usually left out.

Practise next

• Exemplar Multiple Choice Questions Q1–10 — the same formulae in one-step form, useful for checking the sign convention quickly.
• Exemplar Multiple Choice Questions Q11–19 — the three ray-diagram questions, which test the constructions drawn here in recognition form.
• NCERT textbook Chapter 9 — Light: Reflection and Refraction for the end-of-chapter numericals on the mirror and lens formulae.
• Textbook activities for the optical-bench work of Question 38 and the glass-slab tracing of Questions 21 and 32.