Menu Top
Latest Science NCERT Notes and Solutions (Class 6th to 10th)
6th 7th 8th 9th 10th
Latest Science NCERT Notes and Solutions (Class 11th)
Physics Chemistry Biology
Latest Science NCERT Notes and Solutions (Class 12th)
Physics Chemistry Biology

Class 10th Chapters
1. Chemical Reactions And Equations 2. Acids, Bases And Salts 3. Metals And Non-Metals
4. Carbon And Its Compounds 5. Life Processes 6. Control And Coordination
7. How Do Organisms Reproduce? 8. Heredity 9. Light – Reflection And Refraction
10. The Human Eye And The Colourful World 11. Electricity 12. Magnetic Effects Of Electric Current
13. Our Environment



Chapter 9 Light – Reflection And Refraction



Light is a form of energy that enables us to see the world around us. We see objects because they reflect light that falls on them, and this reflected light enters our eyes. For transparent objects, light passes through them (transmission), allowing us to see through them. Many fascinating phenomena, like image formation, twinkling stars, rainbows, and the bending of light, are associated with light.

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. However, when objects are very small, light can bend around them (diffraction), indicating its wave nature. Modern quantum theory reconciles the wave and particle nature of light.

This chapter explores the phenomena of **reflection** and **refraction** of light, particularly focusing on how spherical mirrors and lenses form images, using the simplifying assumption that light travels in straight lines.

Reflection Of Light

**Reflection** is the phenomenon of light bouncing back into the same medium after striking a surface. Highly polished surfaces, like mirrors, are good reflectors of light. Reflection of light follows specific 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 to the reflecting surface at the point of incidence, and the reflected ray all lie in the same plane.

These laws apply to all types of reflecting surfaces, including plane mirrors and spherical mirrors.

You are already familiar with image formation by a **plane mirror**: the image is always virtual (cannot be formed on a screen), erect (upright), the same size as the object, located as far behind the mirror as the object is in front, and laterally inverted (left-right reversal).



Spherical Mirrors

**Spherical mirrors** are mirrors whose reflecting surface is part of a sphere. They can be curved inwards or outwards.

Schematic diagrams showing a concave mirror (reflecting surface curved inwards) and a convex mirror (reflecting surface curved outwards). The non-reflecting back is shaded.

Terms related to spherical mirrors:

Ray diagrams showing parallel rays converging at the principal focus F of a concave mirror and appearing to diverge from the principal focus F of a convex mirror after reflection.

For spherical mirrors with small apertures, the radius of curvature R is twice the focal length f, i.e., **R = 2f**. This means the principal focus F is midway between the pole P and the center of curvature C.


Image Formation By Spherical Mirrors

Spherical mirrors can form various types of images depending on the position of the object relative to the mirror. The image can be real or virtual, erect or inverted, and magnified, diminished, or the same size.

**Image Formation by Concave Mirror:**

The nature, position, and size of the image formed by a concave mirror vary significantly with the object's position. A summary is given in Table 9.1 (from text). Key points:

Summary of image formation by a concave mirror:

Position of the object Position of the image Size of the image Nature of the image
At infinityAt the focus FHighly diminished, point-sizedReal and inverted
Beyond CBetween F and CDiminishedReal and inverted
At CAt CSame sizeReal and inverted
Between C and FBeyond CEnlargedReal and inverted
At FAt infinityHighly enlargedReal and inverted
Between P and FBehind the mirrorEnlargedVirtual and erect

Uses of concave mirrors:

**Image Formation by Convex Mirror:**

Convex mirrors always form **virtual, erect, and diminished** images, regardless of the object's position. The image is always formed behind the mirror, between the pole (P) and the principal focus (F).

Summary of image formation by a convex mirror:

Position of the object Position of the image Size of the image Nature of the image
At infinityAt the focus F, behind the mirrorHighly diminished, point-sizedVirtual and erect
Between infinity and the pole P of the mirrorBetween P and F, behind the mirrorDiminishedVirtual and erect

Uses of convex mirrors:

Question 1. Define the principal focus of a concave mirror.

Answer:

The principal focus of a concave mirror is a point on its principal axis where rays of light that are parallel to the principal axis converge after reflection from the mirror.

Question 2. The radius of curvature of a spherical mirror is 20 cm. What is its focal length?

Answer:

For spherical mirrors of small aperture, the focal length (f) is equal to half of its radius of curvature (R). The relationship is R = 2f, or f = R/2.

Given Radius of curvature, R = 20 cm.

Focal length, f = R/2 = 20 cm / 2 = 10 cm.

The focal length of the spherical mirror is 10 cm.

Question 3. Name a mirror that can give an erect and enlarged image of an object.

Answer:

A **concave mirror** can give an erect and enlarged image when the object is placed between the pole (P) and the principal focus (F) of the mirror. (A convex mirror always gives an erect but diminished image; a plane mirror gives an erect and same-sized image).

Question 4. Why do we prefer a convex mirror as a rear-view mirror in vehicles?

Answer:

We prefer a convex mirror as a rear-view mirror in vehicles for two main reasons:

  1. They always form an **erect image**. This is important for the driver to easily judge the position of vehicles behind.
  2. They provide a **wider field of view**. Being curved outwards, they allow the driver to see a much larger area of traffic behind them compared to a plane mirror. Although the image is diminished (smaller), this wider field of view is crucial for safe driving.


Representation Of Images Formed By Spherical Mirrors Using Ray Diagrams

Ray diagrams are useful for visualizing how spherical mirrors form images and determining the image's nature, position, and size. To draw ray diagrams, we consider the path of a few specific rays originating from the object, whose behaviour after reflection is known. The intersection of at least two reflected rays locates the image.

Convenient rays to consider:

  1. A ray parallel to the principal axis: After reflection, it passes through the principal focus (concave mirror) or appears to diverge from the principal focus (convex mirror).
Ray diagram showing a ray parallel to the principal axis reflecting through F for a concave mirror and reflecting as if coming from F for a convex mirror.

  1. A ray passing through or directed towards the principal focus: After reflection, it emerges parallel to the principal axis.
Ray diagram showing a ray through F reflecting parallel for a concave mirror and a ray directed towards F reflecting parallel for a convex mirror.

  1. A ray passing through or directed towards the centre of curvature (C): After reflection, it is reflected back along the same path because it strikes the mirror normally (perpendicularly).
Ray diagram showing a ray through C reflecting back along C for a concave mirror and a ray directed towards C reflecting back along C for a convex mirror.

  1. A ray incident obliquely at the pole (P): It is reflected obliquely, following the laws of reflection (angle of incidence = angle of reflection with respect to the principal axis as the normal at P).
Ray diagram showing a ray incident at the pole P reflecting at an equal angle to the principal axis.
Ray diagrams showing image formation by a concave mirror for various object positions.
Ray diagrams showing image formation by a convex mirror for various object positions.

By drawing these diagrams, we can visually confirm the results summarized in the tables for concave and convex mirrors.


Sign Convention For Reflection By Spherical Mirrors

To describe image formation mathematically and solve problems, a consistent set of rules for assigning signs (positive or negative) to distances is needed. The **New Cartesian Sign Convention** is used for spherical mirrors:

  1. The **pole (P)** of the mirror is taken as the **origin** (0,0) of the coordinate system.
  2. The **principal axis** is taken as the **x-axis**.
  3. The **object is always placed to the left** of the mirror. Light from the object falls on the mirror from the left.
  4. All distances are measured **from the pole** (P).
  5. Distances measured in the direction of incident light (to the **right** along the principal axis) are taken as **positive (+)**.
  6. Distances measured in the direction opposite to the direction of incident light (to the **left** along the principal axis) are taken as **negative (-)**.
  7. Distances measured perpendicular to and **above** the principal axis (height upwards) are taken as **positive (+)**.
  8. Distances measured perpendicular to and **below** the principal axis (height downwards) are taken as **negative (-)**.
Diagram illustrating the New Cartesian Sign Convention for spherical mirrors, showing the pole as origin, principal axis as x-axis, positive/negative directions for distances and heights.

According to this convention:


Mirror Formula And Magnification

The relationship between object distance (u), image distance (v), and focal length (f) for a spherical mirror is given by the **mirror formula**:

$$ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} $$

This formula is valid for all spherical mirrors and all object positions. Numerical values for u, v, f must be substituted with their proper signs according to the New Cartesian Sign Convention.

**Magnification (m)** produced by a spherical mirror indicates 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).

$$ m = \frac{\text{Height of the image} (h')}{\text{Height of the object} (h)} = \frac{h'}{h} $$

Magnification is also related to the object and image distances:

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

Combining these, $m = \frac{h'}{h} = -\frac{v}{u}$.

Interpretation of magnification sign:

Example 9.1. 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:

Convex mirror, Radius of curvature, R = +3.00 m (positive for convex mirror)

Object distance, u = -5.00 m (object is always on the left)

Find image position (v), nature, and size (magnification m).

Focal length, f = R/2 = (+3.00 m) / 2 = +1.50 m (positive for convex mirror).

Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$:

$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{+1.50 \text{ m}} - \frac{1}{-5.00 \text{ m}} = \frac{1}{1.50} + \frac{1}{5.00}$

$\frac{1}{v} = \frac{1}{1.5} + \frac{1}{5} = \frac{10}{15} + \frac{3}{15} = \frac{13}{15}$

$v = \frac{15}{13} \text{ m} \approx +1.15 \text{ m}$.

The positive sign for v indicates that the image is formed behind the mirror.

Position of the image: The image is formed at a distance of 1.15 m behind the mirror.

Magnification, $m = -\frac{v}{u} = - \frac{+1.15 \text{ m}}{-5.00 \text{ m}} = +0.23$.

Nature of the image: The positive sign of m indicates that the image is **virtual** and **erect**. Since $|m| = 0.23 < 1$, the image is **diminished** (smaller than the object) by a factor of 0.23.

Example 9.2. 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:

Object size, h = +4.0 cm (upright object)

Object distance, u = -25.0 cm (in front of the mirror)

Concave mirror, Focal length, f = -15.0 cm (negative for concave mirror)

Find image distance (v), nature, and size (h'). Screen must be placed at image distance v to get a sharp image.

Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$:

$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-15.0 \text{ cm}} - \frac{1}{-25.0 \text{ cm}} = \frac{1}{-15} + \frac{1}{25}$

$\frac{1}{v} = \frac{-5}{75} + \frac{3}{75} = \frac{-5+3}{75} = \frac{-2}{75}$

$v = -\frac{75}{2} \text{ cm} = -37.5 \text{ cm}$.

The negative sign for v indicates that the image is formed in front of the mirror.

Position of the image: The screen should be placed at a distance of 37.5 cm in front of the mirror to obtain a sharp image.

Nature of the image: Since the image is formed in front of the mirror (v is negative), the image is **real**. Real images formed by mirrors are always **inverted**. We can confirm this from magnification.

Magnification, $m = -\frac{v}{u} = - \frac{-37.5 \text{ cm}}{-25.0 \text{ cm}} = -1.5$.

The negative sign of m confirms that the image is inverted and real. Since $|m| = 1.5 > 1$, the image is **enlarged**.

Size of the image, $h' = m \times h = (-1.5) \times (+4.0 \text{ cm}) = -6.0 \text{ cm}$.

The negative sign of h' indicates the image is inverted (formed below the principal axis). The size of the image is 6.0 cm.

Question 1. Find the focal length of a convex mirror whose radius of curvature is 32 cm.

Answer:

For a convex mirror, the radius of curvature (R) and focal length (f) are taken as positive according to the sign convention.

Given Radius of curvature, R = +32 cm.

Focal length, f = R/2 = +32 cm / 2 = +16 cm.

The focal length of the convex mirror is 16 cm.

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:

Given:

Concave mirror.

Magnification, m = -3 (since the image is real, magnification is negative).

Object distance, u = -10 cm (in front of the mirror).

Find image location (v).

Using the magnification formula $m = -\frac{v}{u}$:

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

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

$v = -3 \times 10 \text{ cm} = -30 \text{ cm}$.

The negative sign for v indicates that the image is formed in front of the mirror.

The image is located at a distance of 30 cm in front of the concave mirror.



Refraction Of Light

**Refraction** is the phenomenon of the change in the direction of light when it passes from one transparent medium to another. This bending of light occurs because the speed of light is different in different media.

Examples of refraction in daily life:

The extent of bending of light depends on the pair of media involved. Light bends at the interface (boundary) separating the two transparent media.


Refraction Through A Rectangular Glass Slab

When a ray of light passes through a rectangular glass slab, it undergoes refraction twice: first at the air-glass interface and second at the glass-air interface.

Consider a ray of light incident obliquely on a rectangular glass slab:

  1. At the first surface (air to glass): Light travels from a rarer medium (air) to a denser medium (glass). The light ray bends **towards the normal**.
  2. Within the glass slab: Light travels in a straight line.
  3. At the second surface (glass to air): Light travels from a denser medium (glass) to a rarer medium (air). The light ray bends **away from the normal**.
Diagram showing a light ray passing through a rectangular glass slab, illustrating bending towards the normal at the first surface and away from the normal at the second surface. The emergent ray is parallel to the incident ray but laterally shifted.

The bending at the two parallel surfaces (air-glass and glass-air) is equal and opposite. As a result, the emergent ray (the ray coming out of the glass slab) is **parallel** to the direction of the incident ray, but it is laterally shifted.

If a light ray is incident normally (perpendicularly) on the interface between two media, it passes straight through without any bending.

Refraction of light obeys specific laws:

  1. The incident ray, the refracted ray, and the normal to the interface of two transparent media at the point of incidence, all lie in the same plane.
  2. For a given pair of media and for light of a specific colour, the ratio of the sine of the angle of incidence ($\angle i$) to the sine of the angle of refraction ($\angle r$) is constant. This is **Snell's Law**:

    $$ \frac{\sin i}{\sin r} = \text{constant} $$

    This constant value is called the **refractive index** of the second medium with respect to the first medium.

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:

When a ray of light travels from air (a rarer medium) into water (a denser medium) obliquely, it **bends towards the normal**. This happens because the speed of light is greater in air than in water. When light enters an optically denser medium from a rarer medium, its speed decreases, causing it to bend towards the normal.

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 × 10⁸ m s⁻¹.

Answer:

Given:

Refractive index of glass with respect to air (approximately vacuum), n$_{glass}$ = 1.50.

Speed of light in vacuum (or air), c = 3 $\times$ 10$^8$ m/s.

The refractive index of a medium is defined as the ratio of the speed of light in vacuum (or air) to the speed of light in the medium:

n$_{glass}$ = $\frac{\text{Speed of light in vacuum (c)}}{\text{Speed of light in glass (v$_{glass}$)}}$

1.50 = $\frac{3 \times 10^8 \text{ m/s}}{v_{glass}}$

$v_{glass} = \frac{3 \times 10^8 \text{ m/s}}{1.50} = 2 \times 10^8 \text{ m/s}$.

The speed of light in the glass is $2 \times 10^8$ m/s.

Question 3. Find out, from Table 9.3, the medium having highest optical density. Also find the medium with lowest optical density.

Answer:

Optical density is related to refractive index; a higher refractive index indicates higher optical density. From Table 9.3 (in the textbook):

  • The medium having the highest refractive index is Diamond (2.42). So, **Diamond** has the highest optical density.
  • The medium having the lowest refractive index is Air (1.0003). So, **Air** has the lowest optical density.

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.

Answer:

The speed of light in a medium is inversely proportional to its refractive index ($v = c/n$). Light travels fastest in the medium with the lowest refractive index.

From Table 9.3 (in the textbook):

  • Refractive index of Kerosene = 1.44
  • Refractive index of Turpentine oil = 1.47
  • Refractive index of Water = 1.33

Comparing the refractive indices, water has the lowest value (1.33). Therefore, light travels fastest in **water** among these three media.

Question 5. The refractive index of diamond is 2.42. What is the meaning of this statement?

Answer:

The refractive index of diamond is 2.42 means that the speed of light in diamond is 2.42 times slower than the speed of light in vacuum (or air, approximately). Specifically, it means that the ratio of the speed of light in vacuum to the speed of light in diamond is 2.42 ($n_{diamond} = c/v_{diamond} = 2.42$). It also indicates that diamond is optically the densest material listed in Table 9.3, causing light to bend significantly towards the normal when entering it from a rarer medium.


The Refractive Index

The **refractive index (n)** quantifies the extent to which a medium can bend light. It is related to the speed of light in different media.

Optical density is related to refractive index. A medium with a higher refractive index is **optically denser** than a medium with a lower refractive index, which is considered **optically rarer**. The speed of light is higher in an optically rarer medium and lower in an optically denser medium. Light bends towards the normal when going from rarer to denser, and away from the normal when going from denser to rarer.


Refraction By Spherical Lenses

A **lens** is a transparent material bounded by two surfaces, at least one of which is spherical. Lenses bend light rays, causing them to converge or diverge.

Diagrams showing a convex lens converging parallel rays and a concave lens diverging parallel rays.

Terms related to spherical lenses:

For thin lenses with small apertures and centers of curvature equidistant from the optical center (symmetrical lenses), there are two principal foci, F$_1$ and F$_2$, located at equal distances from the optical center on opposite sides.


Image Formation By Lenses

Lenses form images by refracting light. The nature, position, and size of the image depend on the type of lens and the position of the object.

**Image Formation by Convex Lens:**

A convex lens can form both real and virtual images, and the image characteristics vary with object position. A summary is given in Table 9.4 (from text).

Summary of image formation by a convex lens:

Position of the object Position of the image Relative size of the image Nature of the image
At infinityAt focus F2Highly diminished, point-sizedReal and inverted
Beyond 2F1Between F2 and 2F2DiminishedReal and inverted
At 2F1At 2F2Same sizeReal and inverted
Between F1 and 2F1Beyond 2F2EnlargedReal and inverted
At focus F1At infinityInfinitely large or highly enlargedReal and inverted
Between focus F1 and optical centre OOn the same side of the lens as the objectEnlargedVirtual and erect

**Image Formation by Concave Lens:**

A concave lens always forms a **virtual, erect, and diminished** image, regardless of the object's position. The image is always formed on the same side of the lens as the object, between the optical centre (O) and the principal focus (F1).

Summary of image formation by a concave lens:

Position of the object Position of the image Relative size of the image Nature of the image
At infinityAt focus F1Highly diminished, point-sizedVirtual and erect
Between infinity and optical centre O of the lensBetween focus F1 and optical centre ODiminishedVirtual and erect

Image Formation In Lenses Using Ray Diagrams

Similar to mirrors, ray diagrams for lenses use specific rays whose paths after refraction are known. Any two of these rays can be used to locate the image.

Convenient rays to consider for lenses:

  1. A ray of light from the object, parallel to the principal axis: After refraction, passes through the principal focus on the other side (convex lens) or appears to diverge from the principal focus on the same side (concave lens).
Ray diagram showing a ray parallel to principal axis refracting through F2 (convex) or appearing to diverge from F1 (concave).

  1. A ray of light passing through or directed towards a principal focus: After refraction, emerges parallel to the principal axis.
Ray diagram showing a ray through F1 refracting parallel (convex) or a ray directed towards F2 refracting parallel (concave).

  1. A ray of light passing through the optical centre (O): It emerges without any deviation.
Ray diagram showing a ray passing undeviated through the optical center O of a lens.

Ray diagrams for convex lenses:

Ray diagrams showing image formation by a convex lens for various object positions.

Ray diagrams for concave lenses:

Ray diagrams showing image formation by a concave lens for various object positions.

Sign Convention For Spherical Lenses

The New Cartesian Sign Convention is used for spherical lenses, similar to mirrors, but with distances measured from the optical centre (O).

  1. The **optical centre (O)** is taken as the **origin**.
  2. The **principal axis** is the **x-axis**.
  3. The **object is always placed to the left** of the lens.
  4. All distances are measured **from the optical centre** (O).
  5. Distances in the direction of incident light (to the **right** along the principal axis) are **positive (+)**.
  6. Distances opposite to incident light (to the **left** along the principal axis) are **negative (-)**.
  7. Heights above the principal axis (upwards) are **positive (+)**.
  8. Heights below the principal axis (downwards) are **negative (-)**.

According to this convention:


Lens Formula And Magnification

The relationship between object distance (u), image distance (v), and focal length (f) for a spherical lens is given by the **lens formula**:

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

This formula is valid for all spherical lenses and object positions, provided distances are used with proper signs.

**Magnification (m)** produced by a lens is the ratio of image height (h') to object height (h):

$$ m = \frac{h'}{h} $$

Magnification is also related to object and image distances:

$$ m = \frac{v}{u} $$

So, $m = \frac{h'}{h} = \frac{v}{u}$.

Interpretation of magnification sign and value is similar to mirrors: positive m for erect/virtual, negative m for inverted/real; $|m|>1$ for magnified, $|m|=1$ for same size, $|m|<1$ for diminished.

Example 9.3. 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:

Concave lens, focal length, f = -15 cm (negative for concave lens).

Image distance, v = -10 cm (concave lens always forms a virtual image on the same side as the object, so v is negative).

Find object distance (u) and magnification (m).

Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$:

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

$\frac{1}{u} = \frac{-3}{30} + \frac{2}{30} = \frac{-3+2}{30} = \frac{-1}{30}$

$u = -30 \text{ cm}$.

The object should be placed at a distance of 30 cm in front of the lens.

Magnification, $m = \frac{v}{u} = \frac{-10 \text{ cm}}{-30 \text{ cm}} = +\frac{1}{3} \approx +0.33$.

The positive sign of m confirms that the image is virtual and erect. The magnification of +1/3 means the image is one-third the size of the object.

Example 9.4. 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:

Object height, h = +2.0 cm (upright object).

Convex lens, focal length, f = +10 cm (positive for convex lens).

Object distance, u = -15 cm (object in front of the lens).

Find image position (v), nature, size (h'), and magnification (m).

Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$:

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

$\frac{1}{v} = \frac{3}{30} - \frac{2}{30} = \frac{3-2}{30} = \frac{1}{30}$

$v = +30 \text{ cm}$.

The positive sign of v indicates that the image is formed on the other side of the lens.

Position of the image: The image is formed at a distance of 30 cm on the other side of the convex lens.

Nature of the image: Since the image is formed on the other side of the lens (v is positive) and the object is real, the image is **real**. Real images formed by lenses are always **inverted**.

Magnification, $m = \frac{v}{u} = \frac{+30 \text{ cm}}{-15 \text{ cm}} = -2$.

The negative sign of m confirms that the image is inverted and real. Since $|m| = 2 > 1$, the image is **enlarged** (two times).

Size of the image, $h' = m \times h = (-2) \times (+2.0 \text{ cm}) = -4.0 \text{ cm}$.

The negative sign of h' indicates the image is inverted (formed below the principal axis). The size of the image is 4.0 cm.

Question 1. Define 1 dioptre of power of a lens.

Answer:

1 Dioptre (1 D) is defined as the power of a lens whose focal length is 1 metre (1 m). P = 1/f, so if f = 1 m, P = 1/1 = 1 D. The unit Dioptre is only used when the focal length is expressed in metres.

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:

Convex lens.

Real and inverted image formed at v = +50 cm (real image formed on the other side, so v is positive).

Image size is equal to object size, so magnification $|m| = 1$. Since the image is real and inverted, m is negative, so m = -1.

Find object position (u) and power of the lens (P).

Using magnification formula $m = \frac{v}{u}$:

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

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

The negative sign for u indicates that the object is placed in front of the lens. So, the needle is placed at a distance of 50 cm in front of the convex lens.

Note: For a convex lens, when the image is real and the same size as the object (m=-1), the object is always placed at 2F, and the image is formed at 2F on the other side. So, u = -2F and v = +2F, and $|u| = |v| = 2f$. Here $|u| = |v| = 50$ cm, so 2f = 50 cm, which means f = 25 cm = 0.25 m. Let's calculate using the lens formula.

Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$:

$\frac{1}{+50 \text{ cm}} - \frac{1}{-50 \text{ cm}} = \frac{1}{f}$

$\frac{1}{50} + \frac{1}{50} = \frac{1}{f}$

$\frac{2}{50} = \frac{1}{f} \implies \frac{1}{25} = \frac{1}{f}$

$f = +25 \text{ cm}$.

The focal length is 25 cm. Convert focal length to metres for power: f = 25 cm = 0.25 m.

Power of the lens, $P = \frac{1}{f (\text{in metres})} = \frac{1}{+0.25 \text{ m}} = +4 \text{ D}$.

The power of the lens is +4 Dioptres.

Question 3. Find the power of a concave lens of focal length 2 m.

Answer:

Given:

Concave lens, focal length, f = 2 m.

For a concave lens, the focal length is negative according to the sign convention. So, f = -2 m.

Power of the lens, $P = \frac{1}{f (\text{in metres})} = \frac{1}{-2 \text{ m}} = -0.5 \text{ D}$.

The power of the concave lens is -0.5 Dioptre.



Intext Questions



Page No. 142

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. 145

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. 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?

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 9.3, 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 9.3.

Answer:

Question 5. The refractive index of diamond is 2.42. What is the meaning of this statement?

Answer:



Page No. 158

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 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: