NCERT Solutions Class 10 Science Chapter 9 In-text Questions — Light – Reflection and Refraction

NCERT Class 10 Science — Light – Reflection and Refraction, the four in-text QUESTIONS boxes on pages 142, 145, 150 and 158. All 14 questions solved.

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

Chapter 9 stops four times to ask whether you have followed the section just finished. The boxes are short, but they are not easy marks: two of them ask for a definition in the exact words the rest of the chapter depends on, and the other two are the first numericals in the chapter where you have to choose the sign yourself rather than copy it from a worked example. Three results carry all fourteen questions:

• Focal length and radius of curvature — $f = \dfrac{R}{2}$ for a spherical mirror.
• Mirror formula and magnification — $\dfrac1v + \dfrac1u = \dfrac1f$, and $m = \dfrac{h’}{h} = -\dfrac{v}{u}$.
• Refractive index and the lens formula — $n = \dfrac{c}{v}$, then $\dfrac1v – \dfrac1u = \dfrac1f$ with $P = \dfrac1f$ in dioptres, $f$ in metres.

Every one of them is used with the New Cartesian sign convention: the object sits to the left, light runs left to right, distances measured with the light are positive and against it negative. So $u$ is always negative, a concave mirror and a concave lens have negative $f$, and a convex mirror and a convex lens have positive $f$.

Key insight. These boxes are where the chapter checks that you can read a description of an image back into signs. “Real and three times magnified” is not a sentence to remember — it is $m = -3$, and once it is written that way the arithmetic is one line. Getting from words to signs is the whole skill; the rest is substitution.

Questions, page 142. The box that follows Section 9.2.2, on the images formed by spherical mirrors.

Page 142, Question 1

Define the principal focus of a concave mirror.

Solution. The definition has to be given in terms of what happens to a particular set of rays, not in terms of where the point happens to sit. Send a beam of light parallel to the principal axis at a concave mirror. Every ray obeys the law of reflection at the point where it strikes, and because the surface curves inward, the reflected rays all cross the principal axis at one point.

That point is the principal focus. It is on the principal axis, it lies in front of the mirror, and it is real — the light actually passes through it, which is why a concave mirror can burn paper held there.

Page 142, Question 2

The radius of curvature of a spherical mirror is 20 cm. What is its focal length?

Solution. The centre of curvature $C$ is the centre of the sphere the mirror is a part of, and the principal focus $F$ sits halfway between the pole and $C$. That is the whole content of the relation $f = R/2$, and it holds for a concave and a convex mirror alike.

$$f = \frac{R}{2} = \frac{20\ \text{cm}}{2} = 10\ \text{cm}$$

The question gives no sign and does not say which kind of mirror it is, so the answer is the size: 10 cm. Had it named a concave mirror, the sign convention would make it $-10$ cm, and for a convex mirror $+10$ cm.

Page 142, Question 3

Name a mirror that can give an erect and enlarged image of an object.

Solution. Work through what each mirror can do. A plane mirror always gives an erect image, but always of the same size, so it is out. A convex mirror always gives an erect image too, but it is always diminished, whatever you do with the object — that is why it can show a whole street in a small mirror.

A concave mirror gives a real, inverted image for most object positions, but not for all. Bring the object closer than the principal focus and the reflected rays diverge; they only appear to come from a point behind the mirror. That image is virtual, erect, and larger than the object. It is the shaving mirror, and the dentist’s mirror.

Page 142, Question 4

Why do we prefer a convex mirror as a rear-view mirror in vehicles?

Solution. Two reasons, and both matter to a driver. First, a convex mirror always forms a virtual, erect image, whatever the distance of the object — so traffic behind appears the right way up, which a concave mirror could not guarantee.

Second, and this is the reason a plane mirror of the same size will not do: because the convex surface curves outward, it gathers rays from a much wider angle. The field of view is far larger, so the driver sees much more of the road behind through the same small piece of glass. The price is that the image is diminished, which is what the warning “objects in mirror are closer than they appear” is about.

Questions, page 145. The box that closes Section 9.2.4, after the mirror formula and magnification have been set up.

Page 145, Question 1

Find the focal length of a convex mirror whose radius of curvature is 32 cm.

Solution. The same relation as before, $f = R/2$, and here the mirror is named, so the sign convention applies. For a convex mirror the centre of curvature and the principal focus both lie behind the mirror — in the direction the light travels after reflection — so $R$ and $f$ are both positive.

$$f = \frac{R}{2} = \frac{+32\ \text{cm}}{2} = +16\ \text{cm}$$

Page 145, Question 2

A concave mirror produces three times magnified (enlarged) real image of an object placed at 10 cm in front of it. Where is the image located?

Solution. Start by turning the words into signs, which is the only step with any thinking in it. The object is 10 cm in front of the mirror, so $u = -10$ cm. The image is real, and a real image formed by a mirror is always inverted, so its magnification is negative: three times enlarged and inverted means $m = -3$, not $+3$.

Now use the magnification relation, which needs no focal length:

$$m = -\frac{v}{u} \quad\Longrightarrow\quad -3 = -\frac{v}{-10\ \text{cm}}$$

$$-3 = \frac{v}{10\ \text{cm}} \quad\Longrightarrow\quad v = -30\ \text{cm}$$

The negative sign puts the image on the same side as the object, in front of the mirror — which is exactly what a real image must be for a mirror, so the answer is consistent with what the question said about it.

Questions, page 150. The box that follows Section 9.3.2, on refractive index, and the two questions in it that send you to Table 9.3.

Page 150, Question 1

A ray of light travelling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal? Why?

Solution. Compare the two media. Water has a refractive index of 1.33 and air very nearly 1, so water is the optically denser of the two. Light travels more slowly in the denser medium — that is what a larger refractive index means, since $n = c/v$.

When a ray crosses obliquely into a medium where it slows down, it bends towards the normal: the angle of refraction is smaller than the angle of incidence. Going the other way, water into air, it would speed up and bend away from the normal.

Page 150, Question 2

Light enters from air to glass having refractive index 1.50. What is the speed of light in the glass? The speed of light in vacuum is $3 \times 10^8\ \text{m s}^{-1}$.

Solution. The refractive index of a medium is the ratio of the speed of light in air (effectively in vacuum) to the speed in that medium:

$$n = \frac{c}{v} \quad\Longrightarrow\quad v = \frac{c}{n}$$

$$v = \frac{3 \times 10^8\ \text{m s}^{-1}}{1.50} = 2 \times 10^8\ \text{m s}^{-1}$$

Page 150, Question 3

Find out, from Table 9.3, the medium having highest optical density. Also find the medium with lowest optical density.

Solution. Optical density is not mass density — the chapter says so directly, and the note under Table 9.3 makes the point with kerosene, which is optically denser than water although it floats on it. Of two media, the one with the larger refractive index is the optically denser.

So this is a matter of reading the table for its largest and smallest entries. Diamond, at 2.42, is the largest in the table. Air, at 1.0003, is the smallest.

Page 150, Question 4

You are given kerosene, turpentine and water. In which of these does the light travel fastest? Use the information given in Table 9.3.

Solution. Since $v = c/n$, the speed of light in a medium is largest where the refractive index is smallest. From Table 9.3: kerosene 1.44, turpentine oil 1.47, water 1.33.

Water has the smallest of the three, so light travels fastest in water. It is worth noticing that this is the opposite of what mass density would suggest — kerosene and turpentine are both less dense than water, and light still travels more slowly in them.

Page 150, Question 5

The refractive index of diamond is 2.42. What is the meaning of this statement?

Solution. The number is a ratio, and saying what it is a ratio of is the whole answer. A refractive index of 2.42 means the speed of light in air (or vacuum) is 2.42 times the speed of light in diamond.

Put the other way round, light in diamond travels at

$$v = \frac{c}{n} = \frac{3 \times 10^8\ \text{m s}^{-1}}{2.42} \approx 1.24 \times 10^8\ \text{m s}^{-1}$$

which is well under half its speed in air. That is also why diamond sparkles: such a large refractive index bends light very sharply and traps much of it inside the stone by total internal reflection.

Questions, page 158. The last box of the chapter, after the power of a lens has been defined.

Page 158, Question 1

Define 1 dioptre of power of a lens.

Solution. Power measures how strongly a lens bends light, and it is defined as the reciprocal of the focal length in metres, $P = 1/f$. A short focal length means sharp bending and so a large power.

The dioptre is the unit that definition produces. Putting $f = 1$ m gives $P = 1$, so one dioptre is the power of a lens whose focal length is one metre.

Page 158, Question 2

A convex lens forms a real and inverted image of a needle at a distance of 50 cm from it. Where is the needle placed in front of the convex lens if the image is equal to the size of the object? Also, find the power of the lens.

Solution. Again, begin by turning the description into signs. The image is real, so for a lens it forms on the far side from the object and $v$ is positive: $v = +50$ cm. It is the same size as the object and inverted, so

$$m = \frac{h’}{h} = -1$$

For a lens, $m = v/u$, so

$$-1 = \frac{+50\ \text{cm}}{u} \quad\Longrightarrow\quad u = -50\ \text{cm}$$

The needle is 50 cm in front of the lens. Now the lens formula gives the focal length:

$$\frac1v – \frac1u = \frac1f \quad\Longrightarrow\quad \frac{1}{50} – \frac{1}{-50} = \frac{2}{50} = \frac{1}{25}$$

$$f = +25\ \text{cm} = +0.25\ \text{m}$$

$$P = \frac1f = \frac{1}{0.25\ \text{m}} = +4\ \text{D}$$

The positive power confirms the lens is convex, as the question said. Notice the pattern worth remembering: an object at $2f$ gives an image at $2f$, the same size and inverted. Here $50 = 2f$, so $f = 25$ cm — the formula was not strictly needed.

Page 158, Question 3

Find the power of a concave lens of focal length 2 m.

Solution. The sign convention does the work. A concave lens is diverging, and its focal point lies on the same side as the incoming light, so its focal length is negative: $f = -2$ m.

$$P = \frac1f = \frac{1}{-2\ \text{m}} = -0.5\ \text{D}$$

The negative sign is not decoration — it is how an optician’s prescription tells a diverging lens from a converging one.

Common mistakes

• Writing $m = +3$ for the page-145 question because the word is “enlarged”. The word that fixes the sign is “real”. A real image in a mirror is always inverted, so $m$ is negative; $m = +3$ would give $v = +30$ cm, an image behind the mirror, which is virtual — contradicting the question.
• Giving $f = 10$ cm for a concave mirror without its sign in the page-142 question. There the mirror is unnamed, so the size alone is the right answer. But the moment a mirror is named, as in the page-145 question, a focal length written without its sign is only half an answer.
• Reading optical density as mass density in the page-150 questions. Kerosene and turpentine are both lighter than water and light still travels more slowly in them. Only the refractive index decides optical density, and the note under Table 9.3 exists to make that point.
• Quoting “refractive index 2.42” back as “diamond is 2.42 times denser”. It is a ratio of speeds, not of densities. The safe form of words is: the speed of light in air is 2.42 times the speed in diamond.
• Forgetting to convert the focal length to metres before taking the power. $f = 25$ cm gives $P = 1/25 = 0.04$ if you leave it in centimetres. The dioptre is defined with $f$ in metres, so it is $1/0.25 = 4$ D.
• Dropping the negative sign on a concave lens’s power. A lens of power $-0.5$ D and one of $+0.5$ D do opposite things to a beam of light, and only the sign distinguishes them.

Practise next

• The end-of-chapter Exercises for this chapter, where Questions 10 to 12 and 14 to 17 use the same two formulae on harder numbers, and Questions 1 to 6 test the same reasoning as a multiple choice.
• Chapter 10, The Human Eye and the Colourful World — its in-text questions apply the lens formula and the power of a lens to the eye’s own defects.