Light – Reflection and RefractionClass 10 Physics Notes

Light - Reflection and Refraction

We can see the world around us because of light. When you are in a dark room, you can't see anything. But as soon as you turn on a light, objects become visible. This happens because objects reflect the light that falls on them. This reflected light enters our eyes, allowing us to see.

Light is also what allows us to see through transparent materials like glass or water, because it can pass, or be transmitted, through them. Light is responsible for many amazing phenomena, such as the formation of images in mirrors, the twinkling of stars, and the beautiful colors of a rainbow.

A key property of light is that it appears to travel in straight lines. We call this the rectilinear propagation of light. You can see this when a small light source casts a sharp shadow of an object. In this chapter, we will use this straight-line path of light, often called a ray of light, to understand how reflection and refraction work.

REFLECTION OF LIGHT

When light hits a highly polished surface, like a mirror, most of it bounces back. This bouncing back of light is called reflection.

Reflection follows two important laws, known as the laws of reflection:

1. The angle of incidence is equal to the angle of reflection.
2. The incident ray, the normal (an imaginary line perpendicular to the surface at the point of incidence), and the reflected ray all lie in the same plane.

These laws apply to all reflecting surfaces, whether they are flat or curved.

Image Formation by a Plane Mirror You are probably most familiar with a flat, or plane mirror. The images formed by a plane mirror have specific properties:

• The image is always virtual (it cannot be projected onto a screen) and erect (upright).
• The size of the image is the same as the size of the object.
• The image is formed as far behind the mirror as the object is in front of it.
• The image is laterally inverted, meaning the left side of the object appears as the right side of the image, and vice versa.

SPHERICAL MIRRORS

A spherical mirror is a mirror whose reflecting surface is curved. It's like a small section cut from a hollow sphere.

There are two main types of spherical mirrors:

• Concave Mirror: The reflecting surface is curved inwards, like the inside of a spoon. It faces towards the center of the sphere it was cut from.
• Convex Mirror: The reflecting surface is curved outwards, like the back of a spoon.

Important Terms for Spherical Mirrors To understand how these mirrors work, we need to know a few key terms:

• Pole (P): The center of the reflecting surface of the mirror. It lies on the surface of the mirror.
• Centre of Curvature (C): The center of the sphere of which the mirror is a part. For a concave mirror, C is in front of it; for a convex mirror, C is behind it.
• Radius of Curvature (R): The radius of the sphere of which the mirror is a part. It is the distance between the pole (P) and the centre of curvature (C).
• Principal Axis: The straight line passing through the pole (P) and the centre of curvature (C).
• Principal Focus (F):
  • For a concave mirror, it is a point on the principal axis where rays of light parallel to the principal axis meet (converge) after reflection.
  • For a convex mirror, it is a point on the principal axis from which parallel rays of light appear to diverge after reflection.
• Focal Length (f): The distance between the pole (P) and the principal focus (F).
• Aperture: The diameter of the reflecting surface of the mirror.

Image Formation by Spherical Mirrors

The type of image a spherical mirror forms depends on the position of the object. The image can be real (can be projected on a screen) or virtual, and it can be enlarged, diminished, or the same size as the object.

Image Formation by a Concave Mirror The nature, position, and size of the image formed by a concave mirror change depending on where the object is placed.

Image formation by a Convex Mirror A convex mirror always forms the same type of image, regardless of the object's position.

Representation of Images Formed by Spherical Mirrors Using Ray Diagrams

To find the position and nature of an image, we can draw ray diagrams. We only need to draw two special rays from the top of the object. The point where these two rays intersect (or appear to intersect) after reflection gives the position of the image.

Here are four principal rays we can use:

1. A ray parallel to the principal axis: After reflection, this ray will pass through the principal focus (F) of a concave mirror or appear to diverge from the principal focus of a convex mirror.
2. A ray passing through the principal focus (F): After reflection, this ray will emerge parallel to the principal axis.
3. A ray passing through the centre of curvature (C): This ray reflects back along the same path because it strikes the mirror at a $90^\circ$ angle (along the normal).
4. A ray incident at the pole (P): This ray is reflected obliquely, making an equal angle with the principal axis (following the laws of reflection).

Uses of concave mirrors

• Torches, search-lights, and vehicle headlights: To produce powerful, parallel beams of light.
• Shaving mirrors: To see a larger, magnified image of the face.
• Dentists' mirrors: To see large images of patients' teeth.
• Solar furnaces: Large concave mirrors are used to concentrate sunlight to produce intense heat.

Uses of convex mirrors

• Rear-view (wing) mirrors in vehicles: They are preferred for this because:
  • They always produce an erect (upright) image.
  • They have a wider field of view than a plane mirror, allowing the driver to see a much larger area of traffic behind them.

Sign Convention for Reflection by Spherical Mirrors

To solve problems involving mirrors, we use a standard set of sign conventions called the New Cartesian Sign Convention.

• The pole (P) of the mirror is taken as the origin.
• The principal axis is the x-axis.

The rules are:

1. The object is always placed to the left of the mirror.
2. All distances are measured from the pole (P).
3. Distances measured to the right of the pole (along the +x-axis) are positive.
4. Distances measured to the left of the pole (along the -x-axis) are negative.
5. Distances measured upwards and perpendicular to the principal axis (along the +y-axis) are positive.
6. Distances measured downwards and perpendicular to the principal axis (along the -y-axis) are negative.

• Object distance ($u$) is always negative.
• Focal length ($f$) of a concave mirror is negative.
• Focal length ($f$) of a convex mirror is positive.

Mirror Formula and Magnification

Mirror Formula This formula relates the object distance ($u$), the image distance ($v$), and the focal length ($f$) of a spherical mirror. $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ This formula is valid for all spherical mirrors in all situations. Remember to use the New Cartesian Sign Convention when substituting values.

Magnification Magnification (m) tells us how much larger or smaller the image is compared to the object. It is the ratio of the height of the image ($h'$) to the height of the object ($h$). $m = \frac{\text{Height of the image } (h')}{\text{Height of the object } (h)} = \frac{h'}{h}$ Magnification is also related to the object distance ($u$) and image distance ($v$): $m = -\frac{v}{u}$

• A negative sign for magnification ($m$) means the image is real and inverted.
• A positive sign for magnification ($m$) means the image is virtual and erect.
• If $|m| > 1$, the image is enlarged.
• If $|m| < 1$, the image is diminished.
• If $|m| = 1$, the image is the same size as the object.

Given

• Radius of curvature, $R = +3.00 \text{ m}$ (Convex mirror, C is behind)
• Object-distance, $u = -5.00 \text{ m}$ (Object is to the left)

To Find

• Image-distance, $v$
• Nature and size of the image (magnification, $m$)

Formula

$f = \frac{R}{2}$ $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ $m = -\frac{v}{u}$

Solution

First, calculate the focal length: $f = \frac{+3.00 \text{ m}}{2} = +1.50 \text{ m}$

Now, use the mirror formula to find $v$: $\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$ $\frac{1}{v} = \frac{1}{+1.50} - \frac{1}{-5.00} = \frac{1}{1.50} + \frac{1}{5.00}$ $\frac{1}{v} = \frac{5.00 + 1.50}{7.50} = \frac{6.50}{7.50}$ $v = \frac{7.50}{6.50} = +1.15 \text{ m}$

The positive sign for $v$ indicates the image is formed 1.15 m behind the mirror.

Now, find the magnification: $m = -\frac{v}{u} = -\frac{+1.15 \text{ m}}{-5.00 \text{ m}} = +0.23$

Final Answer The image is located 1.15 m behind the mirror. The positive sign for $v$ and $m$ indicates that the image is virtual and erect. Since the magnification is 0.23 (less than 1), the image is smaller in size.

Given

• Object-size, $h = +4.0 \text{ cm}$
• Object-distance, $u = -25.0 \text{ cm}$
• Focal length, $f = -15.0 \text{ cm}$ (Concave mirror)

To Find

• Image-distance, $v$
• Image-size, $h'$
• Nature of the image

Formula

$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ $m = \frac{h'}{h} = -\frac{v}{u}$

Solution

First, find the image distance $v$ using the mirror formula: $\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$ $\frac{1}{v} = \frac{1}{-15.0} - \frac{1}{-25.0} = -\frac{1}{15.0} + \frac{1}{25.0}$ $\frac{1}{v} = \frac{-5.0 + 3.0}{75.0} = \frac{-2.0}{75.0}$ $v = -\frac{75.0}{2.0} = -37.5 \text{ cm}$

The screen should be placed at 37.5 cm in front of the mirror. The negative sign for $v$ indicates the image is real.

Now, find the image height $h'$: $h' = -\frac{v \times h}{u} = -\frac{(-37.5 \text{ cm}) \times (+4.0 \text{ cm})}{(-25.0 \text{ cm})}$ $h' = -6.0 \text{ cm}$

Final Answer The screen should be placed 37.5 cm in front of the mirror. The image is real, inverted (indicated by the negative signs for $v$ and $h'$), and enlarged (height is 6.0 cm compared to the object's 4.0 cm).

REFRACTION OF LIGHT

When light travels from one transparent medium to another (like from air to water), it changes direction. This bending of light is called refraction.

Refraction is the reason why:

• The bottom of a swimming pool appears shallower than it really is.
• A pencil partly immersed in water appears bent at the surface.
• Letters under a glass slab appear raised.

This phenomenon occurs because the speed of light is different in different media.

Refraction through a Rectangular Glass Slab When a ray of light passes through a glass slab:

1. It enters from air (a rarer medium) into glass (a denser medium). It bends towards the normal.
2. It then exits from glass back into the air. It bends away from the normal.

The final ray that comes out (the emergent ray) is parallel to the original ray (the incident ray), but it is shifted sideways slightly.

Laws of Refraction Refraction follows two laws:

1. The incident ray, the refracted ray, and the normal to the interface of the two media at the point of incidence, all lie in the same plane.
2. Snell's Law: The ratio of the sine of the angle of incidence ($i$) to the sine of the angle of refraction ($r$) is a constant for a given color of light and for a given pair of media. $\frac{\sin i}{\sin r} = \text{constant}$ This constant is called the refractive index.

The Refractive Index

The refractive index (n) of a medium is a measure of how much the speed of light is reduced inside that medium. It determines how much light will bend when entering that medium.

The refractive index of medium 2 with respect to medium 1 ($n_{21}$) is given by the ratio of the speed of light in medium 1 ($v_1$) to the speed of light in medium 2 ($v_2$): $n_{21} = \frac{\text{Speed of light in medium 1}}{\text{Speed of light in medium 2}} = \frac{v_1}{v_2}$

The absolute refractive index of a medium ($n_m$) is its refractive index with respect to a vacuum (or air, as the speeds are very similar). $n_m = \frac{\text{Speed of light in air } (c)}{\text{Speed of light in the medium } (v)} = \frac{c}{v}$ Light travels fastest in a vacuum, at a speed of $c \approx 3 \times 10^8 \text{ m s}^{-1}$. A medium with a higher refractive index is called optically denser, while one with a lower refractive index is optically rarer.

Refraction by Spherical Lenses

A lens is a piece of transparent material, usually glass, with one or two curved surfaces. Lenses work by refracting light to form images.

There are two main types of spherical lenses:

• Convex Lens (Converging Lens): Thicker in the middle and thinner at the edges. It bends parallel rays of light inward to converge at a single point (the focus).
• Concave Lens (Diverging Lens): Thinner in the middle and thicker at the edges. It spreads parallel rays of light outward, making them appear to diverge from a single point.

Important Terms for Lenses

• Optical Centre (O): The central point of a lens. A ray of light passing through it does not bend.
• Centres of Curvature (C1, C2): A lens has two spherical surfaces, and each is part of a sphere. The centres of these two spheres are the centres of curvature.
• Principal Axis: The imaginary line passing through the two centres of curvature.
• Principal Focus (F): A lens has two principal foci, one on each side.
  • For a convex lens, it's the point where parallel rays converge after refraction.
  • For a concave lens, it's the point from which parallel rays appear to diverge after refraction.
• Focal Length (f): The distance from the optical centre (O) to the principal focus (F).

Image Formation by Lenses

Image Formation by a Convex Lens The image formed by a convex lens depends on the object's position.

Image Formation by a Concave Lens A concave lens always forms a virtual, erect, and diminished image, no matter where the object is placed.

Image Formation in Lenses Using Ray Diagrams

To draw ray diagrams for lenses, we can use any two of these three principal rays:

1. A ray parallel to the principal axis: After refraction, it passes through the principal focus ($F_2$) of a convex lens, or appears to diverge from the principal focus ($F_1$) of a concave lens.
2. A ray passing through the principal focus ($F_1$): After refraction from a convex lens, it emerges parallel to the principal axis.
3. A ray passing through the optical centre (O): It emerges from the lens without any deviation.

Sign Convention for Spherical Lenses

The sign convention for lenses is similar to that for mirrors, with one key difference:

• All distances are measured from the optical centre (O).
• The focal length ($f$) of a convex lens is positive.
• The focal length ($f$) of a concave lens is negative.

Lens Formula and Magnification

Lens Formula The lens formula relates the object distance ($u$), image distance ($v$), and focal length ($f$). $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ Remember to use the correct signs for all quantities when solving problems.

Magnification Magnification ($m$) for a lens is defined in the same way as for a mirror. $m = \frac{\text{Height of the image } (h')}{\text{Height of the object } (h)} = \frac{h'}{h}$ It is also related to $u$ and $v$ by the formula: $m = \frac{v}{u}$ [!note] Notice there is no negative sign in the magnification formula for lenses, unlike the one for mirrors.

• A positive $m$ means the image is virtual and erect.
• A negative $m$ means the image is real and inverted.

Given

• Image-distance, $v = -10 \text{ cm}$ (Concave lens forms virtual image on the same side)
• Focal length, $f = -15 \text{ cm}$ (Concave lens)

To Find

• Object-distance, $u$
• Magnification, $m$

Formula

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ $m = \frac{v}{u}$

Solution

Rearrange the lens formula to solve for $u$: $\frac{1}{u} = \frac{1}{v} - \frac{1}{f}$ $\frac{1}{u} = \frac{1}{-10} - \frac{1}{-15} = -\frac{1}{10} + \frac{1}{15}$ $\frac{1}{u} = \frac{-3 + 2}{30} = \frac{-1}{30}$ $u = -30 \text{ cm}$

The object should be placed 30 cm from the lens.

Now, calculate the magnification: $m = \frac{v}{u} = \frac{-10 \text{ cm}}{-30 \text{ cm}} = \frac{1}{3} \approx +0.33$

Final Answer The object distance is 30 cm. The positive sign for magnification shows the image is erect and virtual. The image is one-third of the size of the object.

Given

• Height of the object, $h = +2.0 \text{ cm}$
• Focal length, $f = +10 \text{ cm}$ (Convex lens)
• Object-distance, $u = -15 \text{ cm}$

To Find

• Image-distance, $v$
• Height of the image, $h'$
• Magnification, $m$
• Nature of the image

Formula

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ $m = \frac{h'}{h} = \frac{v}{u}$

Solution

First, find the image distance $v$: $\frac{1}{v} = \frac{1}{f} + \frac{1}{u}$ $\frac{1}{v} = \frac{1}{10} + \frac{1}{-15} = \frac{1}{10} - \frac{1}{15}$ $\frac{1}{v} = \frac{3 - 2}{30} = \frac{1}{30}$ $v = +30 \text{ cm}$

The positive sign shows the image is formed at 30 cm on the other side of the lens.

Now, calculate the magnification and image height: $m = \frac{v}{u} = \frac{+30 \text{ cm}}{-15 \text{ cm}} = -2$ $h' = m \times h = (-2) \times (+2.0 \text{ cm}) = -4.0 \text{ cm}$

Final Answer The image is formed at a distance of 30 cm on the other side of the lens. The negative signs for $m$ and $h'$ show that the image is real and inverted. The image is 4.0 cm tall and is two times enlarged.

Power of a Lens

The power (P) of a lens is a measure of its ability to converge or diverge light rays. A lens with a short focal length bends light more and is considered more powerful. Power is defined as the reciprocal of the focal length ($f$). $P = \frac{1}{f}$ The SI unit of power is the dioptre (D).

• 1 dioptre is the power of a lens with a focal length of 1 metre ($1 \text{ D} = 1 \text{ m}^{-1}$).
• The power of a convex lens is positive.
• The power of a concave lens is negative.

When multiple lenses are placed in contact, their total power is the simple algebraic sum of their individual powers: $P = P_1 + P_2 + P_3 + \dots$ This is very useful for opticians when combining lenses to find the correct prescription for eyeglasses.