Chapter 9: Light – Reflection and Refraction (Physics)

Rays of light parallel to the principal axis, after reflection from a concave mirror, all meet at a point on the principal axis. This point is the principal focus (F) of the concave mirror.

The point on the principal axis where rays parallel to the axis meet after reflection from a concave mirror.

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

10 cm

A concave mirror, when the object is placed between its pole and principal focus.

A concave mirror (with the object between P and F).

A convex mirror always forms an erect, diminished image, and because it curves outwards it has a much wider field of view than a plane mirror. So the driver can see a large area of traffic behind the vehicle.

It always gives an erect image and has a wide field of view.

$f=\frac{R}{2}=\frac{32}{2}=+16$ cm (positive, as the focus of a convex mirror is behind it).

+16 cm

A real image is inverted, so $m=-3$. With $u=-10$ cm:

$$m=-\frac{v}{u}\Rightarrow -3=-\frac{v}{-10}\Rightarrow v=-30\text{ cm}$$

30 cm in front of the mirror (v = −30 cm).

It bends towards the normal. Water is optically denser than air, so light travels more slowly in water. A ray going obliquely from a rarer to a denser medium slows down and bends towards the normal.

Towards the normal, because light slows down on entering the optically denser water.

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

2 × 10⁸ m/s

Optical density increases with refractive index.

Highest: diamond (n = 2.42). Lowest: air (n = 1.0003).

Refractive indices: water 1.33, kerosene 1.44, turpentine oil 1.47. Since $v=\frac{c}{n}$, light is fastest in the medium with the lowest refractive index.

Water

It means that the speed of light in vacuum is 2.42 times its speed in diamond, i.e. $v_{\text{diamond}}=\frac{c}{2.42}\approx 1.24\times 10^8$ m s⁻¹. Diamond is optically very dense and bends light strongly.

Light travels 2.42 times slower in diamond than in vacuum.

1 dioptre is the power of a lens whose focal length is 1 metre: $1\text{ D}=1{\text{ m}}^{-1}$.

The power of a lens of focal length 1 m.

A convex lens gives a real image of the same size when the object is at $2F$, and the image is also at $2F$ on the other side. So the needle is 50 cm in front of the lens ($u=-50$ cm, $v=+50$ cm).

$$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}=\frac{1}{50}+\frac{1}{50}=\frac{1}{25}\Rightarrow f=25\text{ cm}=0.25\text{ m}$$

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

The needle is 50 cm in front of the lens; P = +4 D.

$P=\frac{1}{f}=\frac{1}{-2}=-0.5$ D

−0.5 D

A lens must be transparent; clay is opaque.

(d) Clay

(d) Between the pole of the mirror and its principal focus.

(b) At twice the focal length (2F₁).

A negative focal length means a concave mirror and a concave (diverging) lens.

(a) both concave.

Both plane and convex mirrors always form erect (virtual) images; a concave mirror gives an inverted image beyond F.

(d) either plane or convex.

A magnifying glass is a convex lens; a shorter focal length gives a larger magnification.

(c) A convex lens of focal length 5 cm.

The object must be placed between the pole and the focus, i.e. at a distance of less than 15 cm (0 < distance < 15 cm). The image is virtual, erect and larger than the object, formed behind the mirror. For example, at 10 cm: $\frac{1}{v}=\frac{1}{-15}-\frac{1}{-10}=\frac{1}{30}$, so $v=+30$ cm and $m=3$.

Less than 15 cm from the mirror; the image is virtual, erect and enlarged.

1. Concave mirror: the bulb is placed at its focus, so the light is reflected as a powerful parallel beam.
2. Convex mirror: it always gives an erect, diminished image and has a wide field of view.
3. Concave mirror: it converges sunlight to its focus, producing a very high temperature there.

(a) Concave (b) Convex (c) Concave; reasons above.

Yes, it forms a complete image, but the image is less bright (dimmer).

Experiment: focus the image of a distant object, or a candle flame, on a screen using a convex lens. Now cover the lower half of the lens with black paper. The image on the screen is still complete, but fainter.

Explanation: rays from every point of the object reach every part of the lens. The uncovered half still refracts rays from all points of the object, so every point of the image is still formed. But only half as much light passes through, so the image is dimmer.

Yes, a complete but dimmer image, since every part of the lens receives light from every point of the object.

$u=-25$ cm, $f=+10$ cm, $h=5$ cm.

$$\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{10}-\frac{1}{25}=\frac{3}{50}\Rightarrow v=\frac{50}{3}\approx 16.7\text{ cm}$$

$$m=\frac{v}{u}=\frac{50/3}{-25}=-\frac{2}{3},h^{\prime }=m\times h=-\frac{2}{3}\times 5\approx -3.3\text{ cm}$$

The image is 16.7 cm behind the lens (on the other side), 3.3 cm tall, real, inverted and diminished.

A concave lens forms a virtual image on the same side as the object: $v=-10$ cm, $f=-15$ cm.

$$\frac{1}{u}=\frac{1}{v}-\frac{1}{f}=-\frac{1}{10}+\frac{1}{15}=-\frac{1}{30}\Rightarrow u=-30\text{ cm}$$

The object is 30 cm in front of the lens.

$u=-10$ cm, $f=+15$ cm.

$$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{15}+\frac{1}{10}=\frac{1}{6}\Rightarrow v=+6\text{ cm}$$

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

The image is 6 cm behind the mirror; it is virtual, erect and diminished (0.6 times).

• $|m|=1$: the image is the same size as the object.
• The positive sign: the image is erect and virtual.

The image is virtual, erect and of the same size as the object.

$f=\frac{R}{2}=+15$ cm, $u=-20$ cm, $h=5$ cm.

$$\frac{1}{v}=\frac{1}{15}+\frac{1}{20}=\frac{7}{60}\Rightarrow v=\frac{60}{7}\approx +8.6\text{ cm}$$

$$m=-\frac{v}{u}=\frac{60/7}{20}=\frac{3}{7},h^{\prime }=\frac{3}{7}\times 5\approx 2.1\text{ cm}$$

The image is 8.6 cm behind the mirror, virtual and erect, about 2.1 cm tall.

$u=-27$ cm, $f=-18$ cm, $h=7$ cm.

$$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=-\frac{1}{18}+\frac{1}{27}=\frac{-3+2}{54}=-\frac{1}{54}\Rightarrow v=-54\text{ cm}$$

$$m=-\frac{v}{u}=-\frac{-54}{-27}=-2,h^{\prime }=-2\times 7=-14\text{ cm}$$

The screen should be 54 cm in front of the mirror; the image is 14 cm tall, real, inverted and enlarged.

$f=\frac{1}{P}=\frac{1}{-2}=-0.5$ m $=-50$ cm

−50 cm; a concave (diverging) lens.

$f=\frac{1}{P}=\frac{1}{1.5}\approx 0.67$ m $\approx 66.7$ cm

About +66.7 cm; a converging (convex) lens.