NCERT Solutions Class 10 Science Chapter 9 — Light – Reflection and Refraction

NCERT Class 10 Science — Light – Reflection and Refraction, Chapter 9 Exercises. All 17 questions solved.

Also here. Solutions to the in-text questions for this chapter and the activities for this chapter are on their own pages.

The exercises at the end of Chapter 9 fall into three groups. Questions 1 to 6 are multiple-choice checks: where an object must stand to give a particular kind of image, and what the sign of a focal length tells you. Questions 7, 8, 9 and 13 ask you to reason about images — which mirror suits which job, what happens when half a lens is covered, and what a magnification of $+1$ means. Questions 10 to 12 and 14 to 17 are numericals. All of them run on three results:

• Mirror formula and magnification — $\dfrac1v + \dfrac1u = \dfrac1f$, with $f = \dfrac R2$, and $m = \dfrac{h’}{h} = -\dfrac vu$.
• Lens formula and magnification — $\dfrac1v – \dfrac1u = \dfrac1f$, and $m = \dfrac{h’}{h} = \dfrac vu$.
• Power of a lens — $P = \dfrac1f$, with $f$ in metres; the unit is the dioptre, $1\ \text{D} = 1\ \text{m}^{-1}$.

Every formula is used with the New Cartesian sign convention. The object is placed to the left, so light travels from left to right. All distances are measured from the pole of the mirror or the optical centre of the lens: distances in the direction the light travels (to the right) are positive, and distances against it (to the left) are negative. Heights above the principal axis are positive and heights below it are negative. So the object distance $u$ is always negative, a concave mirror and a concave lens have a negative focal length, and a convex mirror and a convex lens have a positive one.

Key insight. Let the signs describe the image for you. Put every distance into the formula with its sign, and the signs of the answer tell you everything about the image without a diagram. A negative $v$ for a mirror, or a positive $v$ for a lens, means a real image that can be caught on a screen; the opposite sign means a virtual image. A negative $m$ means inverted and a positive $m$ erect, and a size of $m$ bigger than $1$ means enlarged. Questions 10, 11, 12, 14 and 15 are all read this way, and Question 13 is the same idea asked directly.

Question 1

Which one of the following materials cannot be used to make a lens?

Solution. A lens works by refraction: light has to pass through it, bending as it enters and leaves the curved surfaces. So a lens can only be made from a transparent material. Water, glass and plastic are all transparent — a drop of water acts as a tiny lens, and spectacle lenses are made of glass or plastic. Clay is opaque. Light cannot pass through it at all, so however it is shaped, it cannot refract light to form an image.

Question 2

The image formed by a concave mirror is observed to be virtual, erect and larger than the object. Where should be the position of the object?

Solution. For every object position from far away right up to the focus, a concave mirror makes the reflected rays actually meet in front of it, so the image is real and inverted. Beyond $C$ the image is smaller than the object, at $C$ it is the same size, and between $C$ and $F$ it is larger — but in all three cases it is real and inverted, so options (a), (b) and (c) are ruled out. Only when the object is brought closer than $F$, between $P$ and $F$, do the reflected rays spread out. They never meet in front of the mirror, but they appear to come from a point behind it, and that gives a virtual, erect, enlarged image. This is how a concave shaving mirror is used.

Question 3

Where should an object be placed in front of a convex lens to get a real image of the size of the object?

Solution. An object at $2F_1$, twice the focal length from a convex lens, gives a real, inverted image at $2F_2$ on the other side, of the same size. The lens formula confirms it. With $u = -2f$:

$$\frac1v = \frac1f + \frac1u = \frac1f – \frac1{2f} = \frac1{2f} \quad\Rightarrow\quad v = 2f$$

$$m = \frac vu = \frac{2f}{-2f} = -1$$

A magnification of size $1$ means the same size, and the minus sign means inverted. The other options fail: an object at $F$ sends the image off to infinity, an object at infinity gives a point-sized image at $F_2$, and an object between $O$ and $F$ gives a virtual image.

Question 4

A spherical mirror and a thin spherical lens have each a focal length of $-15\ \text{cm}$. The mirror and the lens are likely to be

Solution. The focal length is measured from the pole or optical centre, and it is negative when the focus lies on the side the light comes from. A concave mirror brings reflected light to a focus in front of it, on the same side as the incoming light, so its $f$ is negative. A convex mirror’s focus is behind it, so its $f$ is positive. A convex lens brings light to a focus on the far side, so its $f$ is positive; a concave lens spreads light out so that it seems to come from a focus on the near side, the side the light comes from, so its $f$ is negative. A negative focal length therefore means a concave mirror and a concave lens.

Question 5

No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be

Solution. A plane mirror always gives an erect, virtual image of the same size, at every distance. A convex mirror always gives an erect, virtual, smaller image, wherever the object is. A concave mirror gives an erect image only when you stand closer than its focus; step back beyond $F$ and your image turns upside down. So a mirror that shows you erect at every distance cannot be concave, and it could be either of the other two. Options (a) and (c) are each only half of the answer.

Question 6

Which of the following lenses would you prefer to use while reading small letters found in a dictionary?

Solution. To read small print we need an enlarged, erect image. A concave lens always gives a smaller image, so (b) and (d) are ruled out. A convex lens gives an enlarged, erect, virtual image when the page is held closer to it than its focal length — this is a magnifying glass.

Between the two convex lenses, the one with the shorter focal length bends light more strongly and magnifies more. To see this, suppose the enlarged image is to be formed $25\ \text{cm}$ from the lens, a comfortable distance for reading, so $v = -25\ \text{cm}$ (virtual, on the same side as the page). For $f = +5\ \text{cm}$ the lens formula gives

$$\frac1u = \frac1v – \frac1f = -\frac1{25} – \frac15 = -\frac6{25} \quad\Rightarrow\quad u = -\frac{25}6\ \text{cm}, \qquad m = \frac vu = \frac{-25}{-25/6} = 6$$

For $f = +50\ \text{cm}$ the same working gives $\dfrac1u = -\dfrac1{25} – \dfrac1{50} = -\dfrac3{50}$, so $u = -\dfrac{50}3\ \text{cm}$ and $m = 1.5$. The $5\ \text{cm}$ lens makes the letters six times as tall; the $50\ \text{cm}$ lens only one and a half times.

Question 7

We wish to obtain an erect image of an object, using a concave mirror of focal length $15\ \text{cm}$. What should be the range of distance of the object from the mirror? What is the nature of the image? Is the image larger or smaller than the object? Draw a ray diagram to show the image formation in this case.

Solution. A concave mirror gives an erect image only when the object is between the pole $P$ and the principal focus $F$, as in Question 2. With the object anywhere beyond $F$ the reflected rays meet in front of the mirror and the image is real and inverted, and with the object exactly at $F$ the reflected rays go off parallel and no image forms at all. So the object must be less than $15\ \text{cm}$ from the mirror.

To check the nature and size, take the object $10\ \text{cm}$ away as an example: $u = -10\ \text{cm}$ and $f = -15\ \text{cm}$.

$$\frac1v = \frac1f – \frac1u = -\frac1{15} + \frac1{10} = \frac{-2+3}{30} = \frac1{30} \quad\Rightarrow\quad v = +30\ \text{cm}$$

$$m = -\frac vu = -\frac{30}{-10} = +3$$

The positive $v$ puts the image $30\ \text{cm}$ behind the mirror, so it is virtual, and the positive $m$ says it is erect and three times as tall. Any other distance under $15\ \text{cm}$ gives the same kind of image; it grows larger as the object moves towards $F$.

Ray diagram. Two rays from the top of the object are enough. A ray parallel to the principal axis reflects through $F$. A ray travelling along the line from $C$ through the top of the object meets the mirror head-on, along a normal, and reflects straight back along itself. The two reflected rays spread apart and never meet in front of the mirror; traced back behind it (dashed lines), they meet at $A’$, $30\ \text{cm}$ behind $P$.

Question 8

Name the type of mirror used in the following situations. (a) Headlights of a car. (b) Side/rear-view mirror of a vehicle. (c) Solar furnace. Support your answer with reason.

Question 8 (a)

Solution. A headlight has to send a strong beam a long way down the road. The bulb is placed at the principal focus of a concave mirror, and a ray passing through the focus of a concave mirror is reflected parallel to the principal axis. So all the light that strikes the mirror leaves as a powerful parallel beam instead of spreading out.

Question 8 (b)

Solution. A convex mirror always gives an erect image, whatever the distance, so the traffic behind is never seen upside down. The image is also smaller than the objects, and because the mirror curves outwards it takes in light from a much wider area than a plane mirror of the same size. The driver therefore sees a large stretch of road behind the vehicle in a small mirror.

Question 8 (c)

Solution. A solar furnace has to make something very hot using sunlight. Rays from the Sun arrive parallel to one another, and a large concave mirror brings parallel rays together at its principal focus. All the sunlight falling on the mirror is concentrated into a small region there, and an object placed at the focus heats up strongly.

Question 9

One-half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object? Verify your answer experimentally. Explain your observations.

Solution. Yes. Each point of the object sends out light in all directions, so rays from every point of the object reach every part of the lens — the upper half and the lower half alike. The rays from one point that pass through any part of the lens are all brought together at the same image point. Covering half the lens stops the rays that fall on that half, but the uncovered half still receives rays from every point of the object and still brings them to their image points. So every part of the image is still formed. What changes is the amount of light: only about half the light now gets through, so the image is less bright.

Experiment. Fix a convex lens on a stand and place a lighted candle some distance in front of it, farther away than the focal length. Move a screen on the other side until a sharp image of the flame appears. Now cover the upper half of the lens with black paper. The image stays in the same place, the same size and complete — the whole flame is still seen, not just half of it — but it becomes fainter. Covering the lower half instead gives the same result. The paper’s shadow does not appear on the screen, because the paper is at the lens and not where the image forms.

Question 10

An object $5\ \text{cm}$ in length is held $25\ \text{cm}$ away from a converging lens of focal length $10\ \text{cm}$. Draw the ray diagram and find the position, size and the nature of the image formed.

Solution. A converging lens is a convex lens, so $f = +10\ \text{cm}$. The object is in front of the lens, so $u = -25\ \text{cm}$, and it stands upright, so $h = +5\ \text{cm}$. The lens formula, rearranged for $v$:

$$\frac1v = \frac1f + \frac1u = \frac1{10} – \frac1{25} = \frac{5-2}{50} = \frac3{50} \quad\Rightarrow\quad v = \frac{50}3 = +16.7\ \text{cm}$$

For a lens the magnification is $m = v/u$:

$$m = \frac vu = \frac{50/3}{-25} = -\frac23, \qquad h’ = m\,h = -\frac23\times5 = -3.3\ \text{cm}$$

The positive $v$ puts the image on the other side of the lens, where the rays really meet, so it is real. The negative $m$ and $h’$ mean it is inverted, and $3.3\ \text{cm}$ is less than $5\ \text{cm}$, so it is smaller. This agrees with the chapter’s table: the object is beyond $2F_1$ (more than $20\ \text{cm}$ away), and the image forms between $F_2$ and $2F_2$.

Ray diagram. A ray from the top of the object parallel to the axis refracts through $F_2$. A ray through the optical centre $O$ goes straight on. A third ray, through $F_1$, emerges parallel to the axis. All three meet at $A’$.

Question 11

A concave lens of focal length $15\ \text{cm}$ forms an image $10\ \text{cm}$ from the lens. How far is the object placed from the lens? Draw the ray diagram.

Solution. A concave lens has a negative focal length, $f = -15\ \text{cm}$. A concave lens always forms a virtual image on the same side as the object, so the image distance is also negative: $v = -10\ \text{cm}$. Rearranging the lens formula for $u$:

$$\frac1u = \frac1v – \frac1f = -\frac1{10} + \frac1{15} = \frac{-3+2}{30} = -\frac1{30} \quad\Rightarrow\quad u = -30\ \text{cm}$$

The object is $30\ \text{cm}$ from the lens, on the same side as the image. The magnification, $m = \dfrac vu = \dfrac{-10}{-30} = +\dfrac13$, shows the image is erect and one-third the height of the object.

Ray diagram. A ray from the top of the object parallel to the axis leaves the lens spreading outwards, as if it had come from $F_1$. A ray through the optical centre goes straight on. Traced back, the spreading ray (dashed) meets the straight ray at $A’$, $10\ \text{cm}$ from the lens.

Question 12

An object is placed at a distance of $10\ \text{cm}$ from a convex mirror of focal length $15\ \text{cm}$. Find the position and nature of the image.

Solution. A convex mirror has its focus behind it, so $f = +15\ \text{cm}$, and the object in front gives $u = -10\ \text{cm}$. From the mirror formula:

$$\frac1v = \frac1f – \frac1u = \frac1{15} + \frac1{10} = \frac{2+3}{30} = \frac16 \quad\Rightarrow\quad v = +6\ \text{cm}$$

$$m = -\frac vu = -\frac{6}{-10} = +0.6$$

The positive $v$ puts the image $6\ \text{cm}$ behind the mirror, where the light does not actually go, so it is virtual. The positive $m$ means erect, and $0.6$ means smaller than the object — exactly what a convex mirror always gives.

Question 13

The magnification produced by a plane mirror is $+1$. What does this mean?

Solution. Magnification is $m = h’/h$. A value of $1$ means $h’ = h$: the image is exactly the same size as the object. The plus sign means $h’$ has the same sign as $h$, so the image is the same way up — erect. For a mirror an erect image is always a virtual one, formed behind the mirror. The mirror formula’s $m = -v/u = 1$ also gives $v = -u$: the image is as far behind the mirror as the object is in front.

Question 14

An object $5.0\ \text{cm}$ in length is placed at a distance of $20\ \text{cm}$ in front of a convex mirror of radius of curvature $30\ \text{cm}$. Find the position of the image, its nature and size.

Solution. The focal length is half the radius of curvature, and it is positive for a convex mirror: $f = +\dfrac{30}2 = +15\ \text{cm}$. The object gives $u = -20\ \text{cm}$ and $h = +5.0\ \text{cm}$.

$$\frac1v = \frac1f – \frac1u = \frac1{15} + \frac1{20} = \frac{4+3}{60} = \frac7{60} \quad\Rightarrow\quad v = \frac{60}7 = +8.57\ \text{cm}$$

$$m = -\frac vu = -\frac{60/7}{-20} = +\frac37 = 0.43, \qquad h’ = m\,h = \frac37\times5.0 = 2.14\ \text{cm}$$

The positive $v$ places the image about $8.6\ \text{cm}$ behind the mirror, so it is virtual; the positive $h’$ means erect; and $2.1\ \text{cm}$ against $5.0\ \text{cm}$ means smaller.

Note on the book’s answer. The key prints the size as $2.2\ \text{cm}$. That comes from rounding the image distance to $8.6\ \text{cm}$ first and then working out $5.0\times8.6/20 = 2.15$. Carried through without rounding, the size is $15/7 = 2.14\ \text{cm}$, which is $2.1\ \text{cm}$ to the same accuracy. The difference is only rounding.

Question 15

An object of size $7.0\ \text{cm}$ is placed at $27\ \text{cm}$ in front of a concave mirror of focal length $18\ \text{cm}$. At what distance from the mirror should a screen be placed, so that a sharp focussed image can be obtained? Find the size and the nature of the image.

Solution. The screen must go where the image forms. For a concave mirror $f = -18\ \text{cm}$; the object gives $u = -27\ \text{cm}$ and $h = +7.0\ \text{cm}$.

$$\frac1v = \frac1f – \frac1u = -\frac1{18} + \frac1{27} = \frac{-3+2}{54} = -\frac1{54} \quad\Rightarrow\quad v = -54\ \text{cm}$$

$$m = -\frac vu = -\frac{-54}{-27} = -2, \qquad h’ = m\,h = -2\times7.0 = -14\ \text{cm}$$

The negative $v$ puts the image $54\ \text{cm}$ in front of the mirror, on the same side as the object, where the reflected rays really meet — so it is real and can be caught on the screen. The negative $h’$ means inverted, and it is twice the size of the object. This fits the chapter’s table: the object is between $C$ ($36\ \text{cm}$) and $F$ ($18\ \text{cm}$), so the image forms beyond $C$ and is enlarged.

Question 16

Find the focal length of a lens of power $-2.0\ \text{D}$. What type of lens is this?

Solution. Power is the reciprocal of the focal length in metres, so the focal length is the reciprocal of the power:

$$f = \frac1P = \frac1{-2.0\ \text{D}} = -0.50\ \text{m} = -50\ \text{cm}$$

A negative focal length (and so a negative power) belongs to a diverging lens, which is a concave lens.

Question 17

A doctor has prescribed a corrective lens of power $+1.5\ \text{D}$. Find the focal length of the lens. Is the prescribed lens diverging or converging?

Solution. Again the focal length is the reciprocal of the power:

$$f = \frac1P = \frac1{+1.5\ \text{D}} = +0.67\ \text{m} \approx +67\ \text{cm}$$

A positive power and focal length belong to a convex lens, which brings light together: it is converging.

Common mistakes

• Question 4: mixing up the sign rules for mirrors and lenses. Students often remember “convex is positive” for lenses and then assume a negative focal length must mean a convex mirror, or a convex lens. The rule is the same for both: the focal length is negative when the focus is on the side the light comes from, and that is true of a concave mirror and a concave lens alike.
• Question 5: choosing “only plane” or “only convex”. Each is true on its own, which is why it is tempting to stop at the first one that comes to mind. But both mirrors always give erect images, so the complete answer is (d).
• Question 7: using the radius instead of the focal length. The object must be inside $F$, at less than $15\ \text{cm}$, not less than $30\ \text{cm}$. An object between $15\ \text{cm}$ and $30\ \text{cm}$ lies between $F$ and $C$ and gives a real, inverted image.
• Question 9: expecting half an image. It feels natural that covering half the lens should cut off half the picture. But every point of the object sends light to the whole lens, so the uncovered half still forms every point of the image; only the brightness drops.
• Question 10: using the mirror formula or the mirror magnification for a lens. Writing $\frac1v + \frac1u = \frac1f$ gives $v = 7.1\ \text{cm}$, which is wrong. A lens needs $\frac1v – \frac1u = \frac1f$ and $m = v/u$ — without the minus sign that the mirror magnification carries.
• Question 11: taking $v = +10\ \text{cm}$. The question just says “10 cm from the lens”, so students put in a positive number. A concave lens always forms its image on the object’s side, so $v = -10\ \text{cm}$. With $+10$ the formula gives $u = +6\ \text{cm}$, an object behind the lens, which makes no sense.
• Questions 12 and 14: giving a convex mirror a negative focal length. A convex mirror’s focus is behind it, so $f$ is positive. In Question 14 there is a second trap: the question gives the radius of curvature, $30\ \text{cm}$, and $f$ is half of it, $15\ \text{cm}$.
• Questions 16 and 17: working in centimetres. $P = 1/f$ gives dioptres only when $f$ is in metres. A power of $-2.0\ \text{D}$ means $f = -0.50\ \text{m}$, not $-0.5\ \text{cm}$.

Practise next

• The in-text questions of this chapter, fourteen across four boxes. Two ask for the definitions the rest of the chapter depends on, and the other two are the first numericals where you have to choose the sign yourself rather than copy it from a worked example.
• Chapter 10, The Human Eye and the Colourful World — the convex lens of the eye, and concave and convex spectacle lenses like those in Questions 16 and 17 used to correct defects of vision.