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Chapter 10: Light – Reflection And Refraction
Light is a form of energy that enables us to see objects. When light falls on an object, the object reflects some of the light. When this reflected light reaches our eyes, we perceive the object. Light can also pass through transparent media, allowing us to see through them. Light exhibits various fascinating phenomena, such as image formation by mirrors and lenses, the bending of light, and the formation of rainbows.
Based on common observations like the formation of sharp shadows by opaque objects, light appears to travel in straight lines. This straight-line path is often represented as a ray of light. While the wave nature and particle nature of light are more complex topics, the phenomena of reflection and refraction can be studied using the simple model of light travelling in straight lines.
10.1 Reflection Of Light
Reflection of light occurs when light bounces off a surface. Highly polished surfaces, like mirrors, are excellent reflectors.
The reflection of light follows two fundamental laws:
- Law 1: The angle of incidence ($\angle i$) is equal to the angle of reflection ($\angle r$).
- Law 2: The incident ray, the normal to the reflecting surface at the point of incidence, and the reflected ray all lie in the same plane.
These laws are valid for all types of reflecting surfaces, including plane mirrors and curved mirrors. Plane mirrors form images that are always virtual and erect, of the same size as the object, located as far behind the mirror as the object is in front, and laterally inverted.
When light reflects off curved surfaces, the images formed can have different properties compared to those from plane mirrors.
10.2 Spherical Mirrors
A spherical mirror is a reflecting surface that forms part of a sphere. Spherical mirrors are classified based on whether their reflecting surface is curved inwards or outwards.
- Concave Mirror: A spherical mirror whose reflecting surface is curved inwards, towards the centre of the sphere. The inner surface is the reflecting surface.
- Convex Mirror: A spherical mirror whose reflecting surface is curved outwards. The outer surface is the reflecting surface.
Key Terms related to Spherical Mirrors:
- Pole (P): The centre point of the reflecting surface of a spherical mirror. It lies on the surface of the mirror.
- Centre of Curvature (C): The centre of the sphere of which the spherical mirror's reflecting surface is a part. It is not part of the mirror surface. For a concave mirror, C is in front; for a convex mirror, C is behind.
- Radius of Curvature (R): The radius of the sphere of which the mirror's reflecting surface is a part. The distance PC is equal to R.
- Principal Axis: An imaginary straight line passing through the pole and the centre of curvature of the spherical mirror. It is normal to the mirror at the pole.
- Principal Focus (F):
- For a concave mirror, rays of light parallel to the principal axis converge to a point on the principal axis after reflection. This point is the principal focus (F).
- For a convex mirror, rays of light parallel to the principal axis diverge after reflection, but appear to come from a point on the principal axis behind the mirror. This point is the principal focus (F).
- Focal Length (f): The distance between the pole (P) and the principal focus (F) of a spherical mirror.
- Aperture: The diameter of the reflecting surface of a spherical mirror (e.g., distance MN in Fig. 10.2).
For spherical mirrors with small apertures, the radius of curvature (R) is related to the focal length (f) by the relation $\textbf{R = 2f}$. This means the principal focus is midway between the pole and the centre of curvature.
10.2.1 Image Formation By Spherical Mirrors
The nature, position, and size of the image formed by a spherical mirror depend on the position of the object relative to the mirror's pole (P), principal focus (F), and centre of curvature (C).
Image Formation by a Concave Mirror: (Table 10.1 summarises observations)
| Position of the object | Position of the image | Size of the image | Nature of the image |
|---|---|---|---|
| At infinity | At the focus F | Highly diminished, point-sized | Real and inverted |
| Beyond C | Between F and C | Diminished | Real and inverted |
| At C | At C | Same size | Real and inverted |
| Between C and F | Beyond C | Enlarged | Real and inverted |
| At F | At infinity | Highly enlarged | Real and inverted |
| Between P and F | Behind the mirror | Enlarged | Virtual and erect |
Concave mirrors can form both real and virtual images, which can be inverted or erect, and diminished, same size, or magnified, depending on the object's position.
Uses of Concave Mirrors: Used in torches, searchlights, and vehicle headlights (to produce parallel beams of light), shaving mirrors (to see enlarged image of face), by dentists (to see large images of teeth), and in solar furnaces (to concentrate sunlight).
10.2.2 Representation Of Images Formed By Spherical Mirrors Using Ray Diagrams
Ray diagrams are useful for visualising image formation by spherical mirrors and determining the position, nature, and size of the image. To draw ray diagrams, we trace the paths of light rays from the object and their reflections from the mirror. The intersection of at least two reflected rays locates the image of a point on the object.
Convenient Rays for Drawing Ray Diagrams:
- A ray parallel to the principal axis:
- Concave mirror: Passes through the principal focus (F) after reflection.
- Convex mirror: Appears to diverge from the principal focus (F) behind the mirror after reflection.
- A ray passing through/directed towards the principal focus (F):
- Concave mirror: Becomes parallel to the principal axis after reflection.
- Convex mirror: Directed towards the principal focus (F) behind the mirror, becomes parallel to the principal axis after reflection.
- A ray passing through/directed towards the centre of curvature (C):
- Concave mirror: Passes through C, is reflected back along the same path.
- Convex mirror: Directed towards C behind the mirror, is reflected back along the same path.
- A ray incident obliquely at the pole (P):
- Incident ray makes an angle with the principal axis at P, reflected ray makes the same angle with the principal axis (following laws of reflection).
Image Formation by a Convex Mirror: Convex mirrors always form virtual, erect, and diminished images, regardless of the object's position.
| Position of the object | Position of the image | Relative size of the image | Nature of the image |
|---|---|---|---|
| At infinity | At the focus F, behind the mirror | Highly diminished, point-sized | Virtual and erect |
| Between infinity and the pole P | Between P and F, behind the mirror | Diminished | Virtual and erect |
Uses of Convex Mirrors: Commonly used as rear-view mirrors in vehicles because they always form erect, diminished images and have a wider field of view due to their outward curvature, allowing the driver to see a larger area of traffic behind them.
10.2.3 Sign Convention For Reflection By Spherical Mirrors
To consistently describe image formation and solve problems involving spherical mirrors, a set of sign conventions, the New Cartesian Sign Convention, is used.
Rules of New Cartesian Sign Convention:
- The pole (P) of the mirror is taken as the origin (0,0).
- The principal axis is taken as the x-axis.
- The object is always placed to the left of the mirror. Incident light travels from left to right.
- All distances are measured from the pole of the mirror.
- Distances measured to the right of the pole (along +x-axis) are taken as positive.
- Distances measured to the left of the pole (along -x-axis) are taken as negative.
- Distances measured perpendicular to and above the principal axis (along +y-axis) are taken as positive (e.g., object height).
- Distances measured perpendicular to and below the principal axis (along -y-axis) are taken as negative (e.g., height of a real, inverted image).
Based on these conventions:
- Focal length (f) of a concave mirror is negative (F is in front, left of P).
- Focal length (f) of a convex mirror is positive (F is behind, right of P).
- Object distance (u) is always negative (object is on the left).
- Image distance (v) is negative for virtual images (formed behind, right of P).
- Image distance (v) is positive for real images (formed in front, left of P).
10.2.4 Mirror Formula And Magnification
The relationship between the object distance ($u$), image distance ($v$), and focal length ($f$) for a spherical mirror is given by the mirror formula.
Mirror Formula:
$$ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} $$This formula is valid for all spherical mirrors (concave and convex) and for all object positions. Proper sign conventions must be used when substituting numerical values into the formula.
The focal length ($f$) is half the radius of curvature ($R$): $f = R/2$. Or $R = 2f$.
Magnification ($m$): Magnification describes how much larger or smaller the image is relative to the object. It is defined as the ratio of the height of the image ($h'$) to the height of the object ($h$).
$$ m = \frac{\text{Height of the image}}{\text{Height of the object}} = \frac{h'}{h} $$Magnification is also related to the object distance ($u$) and image distance ($v$):
$$ m = -\frac{v}{u} $$Interpretation of Magnification Sign:
- If $m$ is positive, the image is virtual and erect (formed above the principal axis, $h'$ is positive).
- If $m$ is negative, the image is real and inverted (formed below the principal axis, $h'$ is negative).
- If $|m| > 1$, the image is magnified (enlarged).
- If $|m| = 1$, the image is the same size as the object.
- If $|m| < 1$, the image is diminished (smaller than the object).
Example 10.1. A convex mirror used for rear-view on an automobile has a radius of curvature of 3.00 m. If a bus is located at 5.00 m from this mirror, find the position, nature and size of the image.
Answer:
Given:
- Mirror type: Convex mirror.
- Radius of curvature, $R = +3.00$ m (positive for convex mirror).
- Object distance, $u = -5.00$ m (object always placed to the left).
Focal length, $f = R/2 = (+3.00 \text{ m}) / 2 = +1.50$ m (positive for convex mirror).
Using the mirror formula, $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$:
$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{+1.50 \text{ m}} - \frac{1}{-5.00 \text{ m}} = \frac{1}{1.50} + \frac{1}{5.00}$
To add fractions, find a common denominator (e.g., 7.50):
$\frac{1}{v} = \frac{5}{7.50} + \frac{1.50}{7.50} = \frac{5 + 1.50}{7.50} = \frac{6.50}{7.50}$
$v = \frac{7.50}{6.50} \text{ m} \approx +1.15 \text{ m}$.
The positive sign of $v$ indicates that the image is formed behind the mirror, at a distance of 1.15 m. This confirms it is a virtual image.
Magnification, $m = -\frac{v}{u} = -\frac{+1.15 \text{ m}}{-5.00 \text{ m}} = +0.23$.
The positive sign of $m$ indicates that the image is erect. The value $|m| = 0.23 < 1$ indicates that the image is diminished (smaller than the object) by a factor of 0.23.
Thus, the image is formed at 1.15 m behind the mirror, is virtual, erect, and diminished.
Example 10.2. An object, 4.0 cm in size, is placed at 25.0 cm in front of a concave mirror of focal length 15.0 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Find the nature and the size of the image.
Answer:
Given:
- Mirror type: Concave mirror.
- Object size (height), $h = +4.0$ cm (above principal axis).
- Object distance, $u = -25.0$ cm (object in front).
- Focal length, $f = -15.0$ cm (negative for concave mirror).
To find the image distance ($v$), use the mirror formula, $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$:
$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-15.0 \text{ cm}} - \frac{1}{-25.0 \text{ cm}} = \frac{1}{-15.0} + \frac{1}{25.0}$
To add fractions, find a common denominator (e.g., 75.0):
$\frac{1}{v} = \frac{-5}{75.0} + \frac{3}{75.0} = \frac{-5 + 3}{75.0} = \frac{-2}{75.0}$
$v = \frac{75.0}{-2} \text{ cm} = -37.5 \text{ cm}$.
The negative sign of $v$ indicates that the image is formed in front of the mirror (on the same side as the object), at a distance of 37.5 cm from the mirror. Since the image is formed in front of the concave mirror, it is a real image. A screen should be placed at 37.5 cm in front of the mirror to obtain the sharp image.
To find the size (height) and nature of the image, calculate magnification, $m = \frac{h'}{h} = -\frac{v}{u}$:
$\frac{h'}{+4.0 \text{ cm}} = -\frac{-37.5 \text{ cm}}{-25.0 \text{ cm}} = - \left(\frac{37.5}{25.0}\right) = -1.5$.
$h' = -1.5 \times (+4.0 \text{ cm}) = -6.0 \text{ cm}$.
The negative sign of $h'$ indicates that the image is inverted. Since $h'$ is negative and $v$ is negative (and the image is in front of the mirror for $v<0$ in concave mirrors), the image is real and inverted.
The value $|m| = 1.5 > 1$ indicates that the image is enlarged (magnified) by a factor of 1.5.
Thus, a real, inverted, and enlarged image of size 6.0 cm is formed at a distance of 37.5 cm in front of the concave mirror.
10.3 Refraction Of Light
Light travels in straight lines within a single transparent medium. However, when light passes from one transparent medium to another (e.g., from air to water or glass), it changes its direction. This phenomenon is called refraction of light.
We observe refraction in everyday life:
- The apparent raising of the bottom of a water tank or pond.
- Letters appearing raised when viewed through a thick glass slab.
- A pencil appearing bent or displaced where it enters the water surface in a glass.
- A lemon in water appearing larger when viewed from the side.
The extent of bending of light when it enters a different medium varies depending on the pair of media. This indicates that light travels at different speeds in different transparent media. Refraction occurs because of the change in the speed of light as it passes from one medium to another.
10.3.1 Refraction Through A Rectangular Glass Slab
Studying the path of light through a rectangular glass slab helps understand refraction.
When a ray of light enters a rectangular glass slab from air obliquely:
- At the first interface (air to glass), the light ray bends towards the normal. This happens because glass is optically denser than air, and the speed of light decreases in a denser medium.
- At the second interface (glass to air), the light ray bends away from the normal. This happens because air is optically rarer than glass, and the speed of light increases in a rarer medium.
For a rectangular glass slab with parallel faces, the emergent ray (exiting the glass) is parallel to the incident ray (entering the glass), but it is slightly shifted sideways (lateral displacement).
If a light ray is incident normally (perpendicularly) on the interface between two media, it passes straight without bending (no deviation).
Refraction of light obeys certain laws:
- Law 1: The incident ray, the refracted ray, and the normal to the interface at the point of incidence all lie in the same plane.
- Law 2 (Snell's Law): For a given pair of media and light of a specific colour, the ratio of the sine of the angle of incidence ($\sin i$) to the sine of the angle of refraction ($\sin r$) is constant. $$ \frac{\sin i}{\sin r} = \text{constant} $$ This constant is called the refractive index of the second medium with respect to the first medium.
10.3.2 The Refractive Index
The refractive index quantifies the extent to which a medium can bend light. It is related to the speed of light in different media.
Light travels fastest in vacuum (speed $c = 3 \times 10^8$ m s⁻¹). Its speed is slightly less in air and significantly less in denser media like glass or water.
The refractive index of medium 2 with respect to medium 1 ($n_{21}$) is defined as the ratio of the speed of light in medium 1 ($v_1$) to the speed of light in medium 2 ($v_2$).
$$ n_{21} = \frac{\text{Speed of light in medium 1}}{\text{Speed of light in medium 2}} = \frac{v_1}{v_2} $$Conversely, the refractive index of medium 1 with respect to medium 2 ($n_{12}$) is $n_{12} = \frac{v_2}{v_1}$. Note that $n_{12} = \frac{1}{n_{21}}$.
The absolute refractive index of a medium ($n_m$) is its refractive index with respect to vacuum or air (as speed in air is very close to vacuum). It is simply called the refractive index of the medium.
$$ n_m = \frac{\text{Speed of light in vacuum/air}}{\text{Speed of light in the medium}} = \frac{c}{v} $$| Material medium | Refractive index | Material medium | Refractive index |
|---|---|---|---|
| Air | 1.0003 | Canada Balsam | 1.53 |
| Ice | 1.31 | Rock salt | 1.54 |
| Water | 1.33 | Carbon disulphide | 1.63 |
| Alcohol | 1.36 | Dense flint glass | 1.65 |
| Kerosene | 1.44 | Ruby | 1.71 |
| Fused quartz | 1.46 | Sapphire | 1.77 |
| Turpentine oil | 1.47 | Diamond | 2.42 |
| Benzene | 1.50 | ||
| Crown glass | 1.52 |
Optical Density: The ability of a medium to refract light is referred to as its optical density. A medium with a higher refractive index is optically denser than a medium with a lower refractive index. Optically denser media have a lower speed of light, and optically rarer media have a higher speed of light. Light bends towards the normal when entering an optically denser medium and away from the normal when entering an optically rarer medium.
10.3.3 Refraction By Spherical Lenses
A lens is a transparent material bounded by one or two spherical surfaces. Lenses bend light rays through refraction and can form images.
- Convex Lens: Thicker in the middle and thinner at the edges. Converges parallel rays of light. Also called a converging lens.
- Concave Lens: Thinner in the middle and thicker at the edges. Diverges parallel rays of light. Also called a diverging lens.
Key Terms related to Lenses:
- Centres of Curvature ($C_1, C_2$): Each spherical surface of a lens is part of a sphere. The centres of these spheres are the centres of curvature of the lens. A lens has two centres of curvature, one on each side.
- Principal Axis: An imaginary straight line passing through the two centres of curvature of the lens.
- Optical Centre (O): The central point of a lens. A ray of light passing through the optical centre goes straight without any deviation.
- Aperture: The effective diameter of the circular outline of a spherical lens.
- Principal Focus (F):
- For a convex lens, rays of light parallel to the principal axis converge to a point on the principal axis on the other side of the lens after refraction. This point is the principal focus (F₂). Similarly, rays passing through F₁ on one side become parallel after refraction. A convex lens has two principal foci, F₁ and F₂.
- For a concave lens, rays of light parallel to the principal axis diverge after refraction and appear to come from a point on the principal axis on the same side of the lens. This point is the principal focus (F₁). Rays directed towards F₂ on the other side become parallel after refraction. A concave lens also has two principal foci, F₁ and F₂.
- Focal Length (f): The distance between the optical centre (O) and a principal focus (F) of a lens. For a convex lens, focal length is usually taken as positive; for a concave lens, it is negative.
10.3.4 Image Formation By Lenses
Lenses form images by bending (refracting) light rays. The nature, position, and size of the image formed by a lens depend on the type of lens and the position of the object relative to the lens.
Image Formation by a Convex Lens: (Table 10.4 summarises observations)
| Position of the object | Position of the image | Relative size of the image | Nature of the image |
|---|---|---|---|
| At infinity | At focus F₂ | Highly diminished, point-sized | Real and inverted |
| Beyond 2F₁ | Between F₂ and 2F₂ | Diminished | Real and inverted |
| At 2F₁ | At 2F₂ | Same size | Real and inverted |
| Between F₁ and 2F₁ | Beyond 2F₂ | Enlarged | Real and inverted |
| At focus F₁ | At infinity | Infinitely large or highly enlarged | Real and inverted |
| Between focus F₁ and optical centre O | On the same side of the lens as the object | Enlarged | Virtual and erect |
A convex lens can form both real and virtual images, which can be inverted or erect, and diminished, same size, or magnified, depending on the object's position. It is a converging lens.
Image Formation by a Concave Lens: (Table 10.5 summarises observations)
| Position of the object | Position of the image | Relative size of the image | Nature of the image |
|---|---|---|---|
| At infinity | At focus F₁, on the same side as object | Highly diminished, point-sized | Virtual and erect |
| Between infinity and optical centre O | Between focus F₁ and optical centre O, on the same side as object | Diminished | Virtual and erect |
A concave lens always forms a virtual, erect, and diminished image, regardless of the object's position. It is a diverging lens.
10.3.5 Image Formation In Lenses Using Ray Diagrams
Ray diagrams help visualise image formation by lenses. Similar to mirrors, we use standard rays whose paths after refraction are known.
Convenient Rays for Drawing Ray Diagrams for Lenses:
- A ray parallel to the principal axis:
- Convex lens: Passes through the principal focus (F₂) on the other side after refraction.
- Concave lens: Appears to diverge from the principal focus (F₁) on the same side after refraction.
- A ray passing through/directed towards the principal focus (F):
- Convex lens: Passing through F₁ on the same side, becomes parallel to the principal axis after refraction.
- Concave lens: Directed towards F₂ on the other side, becomes parallel to the principal axis after refraction.
- A ray passing through the optical centre (O):
- Passes straight through the lens without any deviation.
Using any two of these rays, we can draw ray diagrams to locate the image formed by convex and concave lenses for different object positions.
10.3.6 Sign Convention For Spherical Lenses
The New Cartesian Sign Convention is applied to spherical lenses with measurements taken from the optical centre (O).
Rules of New Cartesian Sign Convention for Lenses:
- The optical centre (O) is taken as the origin (0,0).
- The principal axis is the x-axis.
- The object is always placed to the left of the lens. Incident light travels from left to right.
- All distances are measured from the optical centre.
- Distances measured to the right of the optical centre (along +x-axis) are positive.
- Distances measured to the left of the optical centre (along -x-axis) are negative.
- Distances measured perpendicular to and above the principal axis (along +y-axis) are positive (object height, erect image height).
- Distances measured perpendicular to and below the principal axis (along -y-axis) are negative (inverted image height).
Based on these conventions:
- Focal length (f) of a convex lens is positive (F₂ is on the right).
- Focal length (f) of a concave lens is negative (F₁ is on the left).
- Object distance (u) is always negative.
- Image distance (v) is positive for real images (formed on the right).
- Image distance (v) is negative for virtual images (formed on the left).
10.3.7 Lens Formula And Magnification
The relationship between the object distance ($u$), image distance ($v$), and focal length ($f$) for a spherical lens is given by the lens formula.
Lens Formula:
$$ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} $$This formula is valid for all spherical lenses (convex and concave) and for all object positions. Proper sign conventions must be used when substituting numerical values.
Magnification ($m$): Magnification produced by a lens is the ratio of the height of the image ($h'$) to the height of the object ($h$).
$$ m = \frac{\text{Height of the image}}{\text{Height of the object}} = \frac{h'}{h} $$Magnification is also related to the object distance ($u$) and image distance ($v$):
$$ m = \frac{v}{u} $$Interpretation of Magnification Sign and Value:
- Positive $m$: virtual and erect image.
- Negative $m$: real and inverted image.
- $|m| > 1$: magnified image.
- $|m| = 1$: same size image.
- $|m| < 1$: diminished image.
Example 10.3. A concave lens has focal length of 15 cm. At what distance should the object from the lens be placed so that it forms an image at 10 cm from the lens? Also, find the magnification produced by the lens.
Answer:
Given:
- Lens type: Concave lens.
- Focal length, $f = -15$ cm (negative for concave lens).
- Image distance, $v = -10$ cm (concave lens always forms virtual image on the same side as object, so $v$ is negative).
To find object distance ($u$), use the lens formula, $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$:
$-\frac{1}{u} = \frac{1}{f} - \frac{1}{v} = \frac{1}{-15 \text{ cm}} - \frac{1}{-10 \text{ cm}} = \frac{1}{-15} + \frac{1}{10}$
$-\frac{1}{u} = \frac{-2}{30} + \frac{3}{30} = \frac{-2+3}{30} = \frac{1}{30}$
$\frac{1}{u} = -\frac{1}{30}$, so $u = -30$ cm.
The object should be placed at a distance of 30 cm in front of the lens.
To find magnification ($m$), use $m = \frac{v}{u}$:
$m = \frac{-10 \text{ cm}}{-30 \text{ cm}} = +\frac{1}{3} \approx +0.33$.
The positive sign of $m$ indicates that the image is erect and virtual. The value $|m| < 1$ indicates that the image is diminished (one-third the size of the object).
Example 10.4. A 2.0 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 10 cm. The distance of the object from the lens is 15 cm. Find the nature, position and size of the image. Also find its magnification.
Answer:
Given:
- Lens type: Convex lens.
- Object height, $h = +2.0$ cm.
- Focal length, $f = +10$ cm (positive for convex lens).
- Object distance, $u = -15$ cm (object in front).
To find image distance ($v$), use the lens formula, $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$:
$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{+10 \text{ cm}} + \frac{1}{-15 \text{ cm}} = \frac{1}{10} - \frac{1}{15}$
To subtract fractions, find a common denominator (e.g., 30):
$\frac{1}{v} = \frac{3}{30} - \frac{2}{30} = \frac{3-2}{30} = \frac{1}{30}$
$v = +30$ cm.
The positive sign of $v$ indicates that the image is formed on the other side of the lens (to the right), at a distance of 30 cm. Since the image is formed on the opposite side of the convex lens, it is a real image.
To find image size ($h'$) and magnification ($m$), use $m = \frac{h'}{h} = \frac{v}{u}$:
$\frac{h'}{+2.0 \text{ cm}} = \frac{+30 \text{ cm}}{-15 \text{ cm}} = -2$.
$h' = -2 \times (+2.0 \text{ cm}) = -4.0 \text{ cm}$.
The negative sign of $h'$ indicates that the image is inverted (formed below the principal axis). Since $v$ is positive and the image is inverted, the image is real and inverted.
The value $|m| = 2 > 1$ indicates that the image is enlarged (magnified) by a factor of 2.
Thus, a real, inverted, and enlarged image of size 4.0 cm is formed at a distance of 30 cm on the other side of the lens.
10.3.8 Power Of A Lens
The power ($P$) of a lens is a measure of its ability to converge or diverge light rays. It is defined as the reciprocal of its focal length ($f$).
$$ P = \frac{1}{f} $$
The focal length ($f$) must be expressed in metres (m) for the power to be in the SI unit.
The SI unit of power is dioptre (D). 1 Dioptre is the power of a lens whose focal length is 1 metre. 1 D = 1 m⁻¹.
- The power of a convex lens is positive (because $f$ is positive).
- The power of a concave lens is negative (because $f$ is negative).
Opticians use lens power to prescribe corrective lenses. For example, a prescription of +2.0 D means a convex lens with focal length $f = 1/P = 1/2.0 \text{ m} = +0.5 \text{ m}$. A prescription of -2.5 D means a concave lens with focal length $f = 1/P = 1/(-2.5) \text{ m} = -0.4 \text{ m}$.
When multiple thin lenses are placed in contact, their total power is the algebraic sum of their individual powers: $P_{\text{total}} = P_1 + P_2 + P_3 + ...$ This additive property makes it convenient to combine lenses for specific purposes (e.g., in optical instruments).
Question 1. Define 1 dioptre of power of a lens.
Answer:
1 dioptre (1 D) is the power of a lens whose focal length is 1 metre (1 m).
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.
Answer:
Given:
- Lens type: Convex lens.
- Image is real and inverted.
- Image distance, $v = +50$ cm (real image formed on the opposite side).
- Image size is equal to object size, $h' = -h$ (inverted image).
Magnification, $m = \frac{h'}{h} = \frac{-h}{h} = -1$.
Also, $m = \frac{v}{u}$. So, $-1 = \frac{+50 \text{ cm}}{u}$.
$u = -50$ cm.
The needle (object) is placed at a distance of 50 cm in front of the convex lens.
Now, find the power of the lens. First, find the focal length ($f$). Using the lens formula, $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$:
$\frac{1}{f} = \frac{1}{+50 \text{ cm}} - \frac{1}{-50 \text{ cm}} = \frac{1}{50} + \frac{1}{50} = \frac{2}{50} = \frac{1}{25}$
$f = +25$ cm.
Convert focal length to metres for power calculation: $f = +25 \text{ cm} = +0.25 \text{ m}$.
Power, $P = \frac{1}{f} = \frac{1}{+0.25 \text{ m}} = +4 \text{ m}^{-1} = +4 \text{ D}$.
The power of the convex lens is +4 D.
Question 3. Find the power of a concave lens of focal length 2 m.
Answer:
Given:
- Lens type: Concave lens.
- Focal length, $f = -2$ m (negative for concave lens).
Power, $P = \frac{1}{f} = \frac{1}{-2 \text{ m}} = -0.5 \text{ m}^{-1} = -0.5 \text{ D}$.
The power of the concave lens is -0.5 D.
Intext Questions
Page No. 168
Question 1. Define the principal focus of a concave mirror.
Answer:
Question 2. The radius of curvature of a spherical mirror is 20 cm. What is its focal length?
Answer:
Question 3. Name a mirror that can give an erect and enlarged image of an object.
Answer:
Question 4. Why do we prefer a convex mirror as a rear-view mirror in vehicles?
Answer:
Page No. 171
Question 1. Find the focal length of a convex mirror whose radius of curvature is 32 cm.
Answer:
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?
Answer:
Page No. 176
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?
Answer:
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 $3 \times 10^8 m s^{-1}$.
Answer:
Question 3. Find out, from Table 10.3, the medium having highest optical density. Also find the medium with lowest optical density.
Answer:
Question 4. You are given kerosene, turpentine and water. In which of these does the light travel fastest? Use the information given in Table 10.3.
Answer:
Question 5. The refractive index of diamond is 2.42. What is the meaning of this statement?
Answer:
Page No. 184
Question 1. Define 1 dioptre of power of a lens.
Answer:
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.
Answer:
Question 3. Find the power of a concave lens of focal length 2 m.
Answer:
Exercises
Question 1. Which one of the following materials cannot be used to make a lens?
(a) Water
(b) Glass
(c) Plastic
(d) Clay
Answer:
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?
(a) Between the principal focus and the centre of curvature
(b) At the centre of curvature
(c) Beyond the centre of curvature
(d) Between the pole of the mirror and its principal focus.
Answer:
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?
(a) At the principal focus of the lens
(b) At twice the focal length
(c) At infinity
(d) Between the optical centre of the lens and its principal focus.
Answer:
Question 4. A spherical mirror and a thin spherical lens have each a focal length of –15 cm. The mirror and the lens are likely to be
(a) both concave.
(b) both convex.
(c) the mirror is concave and the lens is convex.
(d) the mirror is convex, but the lens is concave.
Answer:
Question 5. No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be
(a) only plane.
(b) only concave.
(c) only convex.
(d) either plane or convex.
Answer:
Question 6. Which of the following lenses would you prefer to use while reading small letters found in a dictionary?
(a) A convex lens of focal length 50 cm.
(b) A concave lens of focal length 50 cm.
(c) A convex lens of focal length 5 cm.
(d) A concave lens of focal length 5 cm.
Answer:
Question 7. We wish to obtain an erect image of an object, using a concave mirror of focal length 15 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.
Answer:
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.
Answer:
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.
Answer:
Question 10. An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the image formed.
Answer:
Question 11. A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object placed from the lens? Draw the ray diagram.
Answer:
Question 12. An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.
Answer:
Question 13. The magnification produced by a plane mirror is +1. What does this mean?
Answer:
Question 14. An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.
Answer:
Question 15. An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 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.
Answer:
Question 16. Find the focal length of a lens of power – 2.0 D. What type of lens is this?
Answer:
Question 17. A doctor has prescribed a corrective lens of power +1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging?
Answer: