Refraction Through a Lens

A concave lens is thick at its periphery and thin in the middle. In other words, a lens which is bent inwards in the middle, is the concave lens. Such a lens diverges the light rays incident on it, so it is called a diverging lens.

Below diagram shows the refraction of two light rays incident parallel to the principal axis on a convex lens. The convex lens is shown as a combination of a glass slab and two triangular glass prisms:

Below diagram shows the refraction of two light rays incident parallel to the principal axis on a concave lens. The concave lens is shown as a combination of a glass slab and two triangular glass prisms:

Convergent action of Convex lens is shown in the below diagram:

Divergent action of Concave lens is shown in the below diagram:

(a) Below diagram shows the incident ray and its corresponding emergent ray:

(b) The incident ray when produced meets the principal axis at a point F, which is called as the first focus .

(a) Below ray diagram shows the incident ray and its corresponding emergent ray:

(b) The point through which the incident ray passes is known as the first focus.

The point at which a beam of light incident on a convex lens parallel to its principal axis converges is called the second focus. Below ray diagram shows the second focus of a convex lens:

The point at which a beam of light incident on a concave lens parallel to its principal axis converges is called the second focus. Below ray diagram shows the second focus of a concave lens:

(i) Below diagram shows the refraction of a parallel oblique beam of light by a convex lens:

(ii) Below diagram shows the refraction of a parallel oblique beam of light by a concave lens:

(i) The lens formed by the combination is a convex lens.

(ii) The line XX' is called the principal axis.

(iii) Below diagram shows the path of the incident ray AB after passing through the convex lens:

(iv) The point F is called a focal point.

(i) The combination forms a concave lens.

(ii) The line XX' is called the principal axis.

(iii) Below diagram shows the path of the incident ray AB after passing through the concave lens:

(iv) The point F is called a focal point or focus.

(a) Below ray diagram shows the path taken by the light ray AB after it emerges from the thin concave lens:

(b) Below ray diagram shows the path taken by the light ray AB after it emerges from the thin convex lens:

(a) Below ray diagram shows the path taken by the light ray AB after refraction through the thin convex lens:

(b) Below ray diagram shows the path taken by the light ray AB after refraction through the thin concave lens:

concavo-convex

Reason — A concavo convex lens has one surface convex and the other surface concave such that it is thicker in the middle as compared to its periphery.

one surface plane and the other surface convex

Reason — A plano-concave lens has one surface plane and the other surface convex.

a convexo-concave lens is thicker in the middle and has a converging action.

Reason — A convexo-concave lens has one surface concave and the other surface convex such that it is thinner in the middle as compared to its periphery.

downward, upward

Reason — A convex lens in its upper part has a prism with its base downward and a concave lens in its upper part has a prism with its base upward as shown in fig below:

(b) convex lens converges the incident rays towards the principal axis whereas concave lens diverges

Reason — Difference between a convex and concave lens is that a convex lens converges the incident rays towards the principal axis whereas concave lens diverges the incident rays away from the principal axis.

its first focus

Reason — When the incident ray passes through the first focus then it emerges parallel to the principal axis of the lens after refraction. We can understand this better with the help of diagrams given below —

Convex Lens

Concave Lens

its second focus

Reason — A ray of light incident on a lens parallel to its principal axis, after refraction passes through or appears to come from its second focus. We can understand this better with the help of the diagrams given below —

Convex Lens

Concave Lens

the radius of curvature of the two surfaces of the lens are equal.

Reason — The radius of curvature of the two surfaces of the lens are equal in an equi-convex or equi-concave lens.

both (1) and (3)

Reason — The focal length of a lens depends upon two factors:

1. The refractive index of the material of lens relative to its surrounding medium.
2. The radii of curvature of the two surfaces of lens.

increases

Reason — The focal length of a lens depends on the refractive index of the material of lens relative to its surrounding medium. When a lens is placed in water instead of air, its focal length increases.

does not change, decreases

Reason — If part of the lens is covered, its focal length remains unchanged, only the amount of light entering the lens decreases due to which the intensity of image decreases but the position, size and nature of image formed by it do not change.

more

Reason — The focal length of a lens depends on the radii of curvature of the two surfaces of lens and the focal length of a thin convex lens is more than that of a thick lens.

radius of curvature

Reason — The radius of the sphere whose part is the lens surface is called the radius of curvature of that surface.

when we say focal length of a lens, we mean the first focal length of the lens.

Reason — When we refer to the focal length of a lens, it generally implies the magnitude of the focal length, regardless of whether it is the first or second focal length because focal length is the distance from the lens at which parallel rays of light converge or appear to converge after passing through the lens.

Two types of lenses are —

1. Convex or converging lens
2. Concave or diverging lens

(a) The difference between a convex and a concave lens in terms of appearance is as follows —

(b) The difference between a convex and a concave lens in terms of action on the incident light is as follows —

The line joining the centres of curvature of the two surfaces of the lens is called the principal axis.

Optical centre is a point on the principal axis of the lens such that a ray of light passing through this point emerges parallel to its direction of incidence. The optical centre is thus the centre of the lens.

The point O is the optical centre of the convex and concave lenses .

(a) This point on the principal axis is called Optical Centre.

(b) Below labelled diagram shows the Optical Centre for Convex and Concave lens:

A light ray can enter a lens from either side, therefore, a lens has two principal foci, one on either side of the lens. These are known as first focal point( or first focus) F₁ and the second focal point ( or second focus ) F₂

For a convex lens, the first focal point is a point F₁ on the principal axis of the lens such that the rays of light coming from it, after refraction through the lens, become parallel to the principal axis of the lens.

For a convex lens, the second focal point is a point F₂ on the principal axis of the lens such that the rays of light incident parallel to the principal axis, after refraction from the lens, pass through it.

A light ray can enter a lens from either side, therefore, a lens has two principal foci, one on either side of the lens. These are known as first focal point( or first focus) F₁ and the second focal point ( or second focus ) F₂

For a concave lens, the first focal point is a point F₁ on the principal axis of a lens such that the incident rays of light appearing to meet at it, after refraction from the lens become parallel to the principal axis of the lens.

For a concave lens, the second focal point is a point F₂ on the principal axis of the lens such that the rays of light incident parallel to the principal axis, after refraction from the lens, appear to be diverging from this point.

Below diagram represents the second focus of a concave lens:

Below diagram represents the second focus of a convex lens:

A lens is a transparent refracting medium bounded by either two spherical surfaces or one surface spherical and another surface plane.

There are two types of lenses —

1. Convex or converging lens
2. Concave or diverging lens

(i) An equiconvex lens or bi convex has both its surfaces convex, and is a converging lens.

(ii) A concavo-convex lens is thicker in the middle and thinner at the edges and has a converging action on a light beam. Hence, a concavo-convex lens is a converging lens.

A lens is called an equiconvex or equi-concave lens, when the radii of curvature of the two surfaces of lens are equal.

The distance of focus from the optical centre of the lens, is called its focal length.

A lens has two focal length.

1. First focal length — the distance from the optical centre O of the lens to its first focal point F₁ is called the first focal length.
2. Second focal length — the distance from the optical centre O of the lens to its second focal point F₂ is called the second focal length.

A plane normal to the principal axis, passing through the focus, is called the focal plane of a lens.

A lens has two focal plane.

(a) A plane passing through the first focal point and normal to the principal axis of the lens is called the first focal plane.

(a) A plane passing through the second focal point and normal to the principal axis of the lens is called the second focal plane.

(a) When the medium is the same on either side of the lens then it has both its focal lengths equal.

(b) A ray passes undeviated through the lens, when it is incident at the optical centre of the lens.

(a) If half part of a convex lens is covered, the focal length does not change, but the intensity of the image decreases.

(b) A convex lens is placed in water. Its focal length will increase.

(c) The focal length of a thin convex lens is more than that of a thick convex lens.

(a) LL' is a convex lens.

(b) The points O and O’ are called as first and second focal points.

(c) Completed diagram showing the formation of the image of the object AB is given below:

(d) The three characteristics of the image are magnified, virtual and upright.

(e) This action of lens is used in the magnifying glass.

(i) LL' is a concave lens.

(ii) The points O and O' are called the second and first focal point.

(iii) Below is the completed diagram showing the image of the object AB:

(iv) The three characteristics of the image are virtual, erect and diminished.

(a) Below is the completed diagram showing the position of the image CD and the position of eye from where the image can be viewed:

(b) The three characteristics of the image are magnified, virtual and upright.

(i) Below ray diagram shows the position of the image:

(ii) Position of images will be more than twice the focal length of the lens.

(iii) The three characteristics of this image are magnified, real and inverted.

(iv) When the object move towards F₁, the image will shift away from F₂ and it will be magnified.

The image will form at infinity and it is highly magnified at F₁.

Between F₁ and optical centre, the image will form on the same side of the object and will be magnified.

(a) The object is placed between F₁ and 2F₁.

(b) Below is the ray diagram showing an object placed between F₁ and 2F₁ of the lens and its image is formed beyond 2F² of the lens:

(c) The three characteristics of the image is real, inverted and magnified.

(a) The object is placed at a distance equal to twice the focal length of the lens (i.e., Object is placed at 2F₁).

(b) Below diagram illustrates the formation of an image equal in size of the object by a convex lens:

(c) The image formed is real and inverted

(a) When a lens forms an erect, magnified and virtual image of an object then we can say that it is a convex lens.

(b) The object is placed between the lens and focus (F₁)

(c) Below ray diagram shows the formation of the image:

(d) This principle is used in magnifying glass.

(a) When a lens forms an image between the object and the lens then we can say that it is a concave lens.

(b) Below ray diagram shows the formation of the image:

(c) The three characteristics of the image are virtual, erect and diminished.

When the candle is placed beyond 2F₁, the image is formed between F₂ and 2F₂.

The image is real, diminished and inverted.

In the ray diagram shown below, the candle is represented as AB and its real, inverted and diminished image is represented as A'B' formed between F₂ and 2F₂.

When the object is at any finite distance from the concave lens, the image formed is virtual, erect and diminished. Hence, a diverging lens (concave lens) cannot form a real image.

When the object is placed between 2F₁ and F₁ then the image is formed beyond 2F₂.

The image is real, enlarged and inverted.

It is a concave lens.

Below ray diagram shows the formation of an image by a concave lens:

When the object is placed between the convex lens and focal point F₁ then the image is formed on the same side of the lens. The image so formed is enlarged.

As the image formed is enlarged so we can say that the converging lens can be used as a magnifier.

Below ray diagram shows how a converging lens can form an image of the Sun:

The Sun is at infinity so convex lens forms its image at the second focal point which is real and very much diminished in size.

The rays of light from the sun converge to a single point called the focus of the lens. This concentrated image of the Sun burns the paper kept below the lens. Hence, the term burning glass is used for the lens.

(i) When the object is situated at infinity, the position of the image is at F₂. The image is real and inverted. It is highly diminished in size.

(ii) When the object is situated beyond 2F₁, the position of the image is between F₂ and 2F₂. The image is real and inverted. It is diminished in size.

(iii) When the object is situated at 2F₁, the position of the image is at 2F₂. The image is real and inverted. It is of the same size as the object.

(iv) When the object is situated between 2F₁ and F₁, the position of the image is beyond 2F₂. The image is real and inverted. It is magnified in size.

(v) When the object is situated at F₁, the position of the image is at infinity. The image is real and inverted. It is highly magnified in size.

(vi) When the object is situated between lens and F₁, the position of the image is on the same side . The image is virtual and upright. It is magnified in size.

(i) When the object is at infinity, the image is formed at the second focus F₂, on the side of the object. It is virtual and erect. It is highly diminished in size.

(ii) When the object is at finite distance from the concave lens, the image is formed between the focus and optical centre, on the same side of lens as the object. It is virtual and erect. It is diminished in size.

appears to come from the second focus

Reason — For a concave lens, the second focal point is a point F₂ on the principal axis of the lens such that the rays of light incident parallel to the principal axis, after refraction from the lens, appear to be diverging from this point.

Reason — For a convex lens, the first focal point is a point F₁ on the principal axis of the lens such that the rays of light coming from it, after refraction through the lens, become parallel to the principal axis of the lens.

In option 2, A ray of light incident parallel to the principal axis of the lens, after refraction appears to come from the second focus F₂ and not F₁.

In option 3, A ray of light incident parallel to the principal axis of the lens, after refraction passes through the second focus F₂ and not F₁

In option 4, A ray of light directed towards the first focus F₁ and not F₂, emerges parallel to the principal axis after refraction.

Hence, fig 1 is correct.

A virtual image can be obtained on the screen.

Reason — A virtual image cannot be obtained on a screen because they are formed by the apparent intersection of rays rather than the actual convergence of rays.

terrestrial telescope

Reason — The image formed in the above case will be real, inverted and of the same size as object hence, it is used in terrestrial telescope for erecting the inverted images formed by the objective lens.

10 cm

Reason —

Given, object distance = image distance.

u = 20 cm

Hence, object must be at 2f.

Therefore,

$u = 2f \\[0.5em]$

Substituting the value of u in the equation above, we get,

$20 = 2f \\[0.5em] \Rightarrow f = \dfrac{20}{2} \\[0.5em] \Rightarrow f = 10 \text{ cm} \\[0.5em]$

virtual and enlarged

Reason — Whenever, the object is placed between the optical centre and focus of a convex lens, the image is virtual, erect and magnified.

We can understand this better with the help of the diagram given below.

virtual, upright, and diminished

Reason — Irrespective of the position of the object, a concave lens always forms an image which is virtual, upright and diminished and it is situated on the side of the object between the focus and the lens.

2F₂

Reason — For a convex lens when the object is at 2F₁, then image is formed at 2F₂ as shown in fig below

in Galilean telescope

Reason — A concave lens is used in a Galilean telescope, as the image formed is virtual, erect and highly diminished.

mid point between the optical centre and second focus of the lens

Reason — For a concave lens, when the object is at a distance equal to the focal length of the lens, the image is formed at mid point between the optical centre and second focus of the lens.

assertion is false but reason is true.

Explanation

Assertion (A) is false. The focal length of a lens does change when different colours of light are used. Different colours have different wavelength so focal length is different.

Reason (R) is true. The focal length (f) of a lens does depend on the refractive index of the material of the lens. It is inversely proportional to the refractive index of the medium μ.

(i) A ray of light incident at the optical centre O of the lens passes undeviated through the lens.

Convex Lens

Concave Lens

(ii) A ray of light incident parallel to the principal axis of the lens, after refraction passes through the second focus F₂ (in a convex lens) or appears to come from the second focus F₂ (in a concave lens), respectively.

Convex Lens

Concave Lens

(iii) A ray of light passing through the first focus F₁ (in a convex lens) or directed towards the first focus F₁ (in a concave lens), emerges parallel to the principal axis after refraction, respectively.

Convex Lens

Concave Lens

Completed diagrams showing the path of the rays A and B as they emerge out of the lens are given below:

Convex Lens

Concave Lens

(a) The lens is a convex lens. A convex lens forms an upright and magnified image, when the position of the object is between the lens and F₁.

(b) Below labelled ray diagram shows the image formation:

(a) In order to form the image at infinity, the object should be placed at focus, in front of a convex lens.

(b) In order to form the image of same size as the object, the object should be placed at 2F, in front of a convex lens.

(c) In order to form an inverted and enlarged image, the object should be placed between F and 2F, in front of a convex lens.

(d) In order to form an upright and enlarged image, the object should be placed between optical centre and focus, in front of a convex lens.

(a) It is a convex lens as a convex lens forms an inverted image when the object is placed anywhere from infinity to F₁.

(b) The image formed by a convex lens is real.

(a) A concave lens always forms an erect and virtual image.

(b) The image formed by a concave lens is always diminished.

No, a concave lens cannot form an image of size of two times that of the object as a concave lens diverges the rays incident on it, and the image is always diminished.

A concave lens diverges the rays incident on it, and so the image formed by a concave lens is always virtual and diminished.

The virtual image formed by a convex lens will be erect and magnified.

(a) An object is placed at a distance of more than 40 cm from a convex lens of focal length 20 cm. The image formed is real, inverted and diminished.

(b) An object is placed at a distance 2f from a convex lens of focal length f. The image formed is equal to that of the object.

(c) An object is placed at a distance 5 cm from a convex lens of focal length 10 cm. The image formed is virtual, upright and magnified.

(a) False
Reason — Convex lenses have a convergent action, not a divergent action. They converge parallel incident rays of light to a focal point after refraction, hence they are often called converging lenses.
Concave lenses, on the other hand, have a divergent action. They diverge parallel incident rays of light as if they are coming from a virtual focus point, hence they are often called diverging lenses.

(b) False
Reason — Real images are formed by convex lenses, not concave lenses. Concave lenses are used to create virtual images that are upright and diminished.

(c) False
Reason — A ray of light incident at the optical centre O of a lens, passes undeviated through the lens after refraction.

(d) True
Reason — At the optical center, the incident ray passes directly through without any change in direction. Hence, it emerges from the lens with the same direction as it had when entering the lens.

(e) False
Reason — A concave lens always forms a diminished, virtual image, regardless of the distance of the object from it.

In order to obtain a parallel beam of light, the source of light should be placed at the first focal point i.e. the focal point on the left of the optical centre of the convex lens.

f = $\dfrac{uv}{u-v}$

Reason — The correct lens formula is f = $\dfrac{uv}{u-v}$

For a convex lens, v is positive for a virtual image and negative for a real image.

Reason — For a convex lens, v is positive for a real image and negative for a virtual image.

both positive as well as negative

Reason — The magnification m is:

• Positive when the image is upright (virtual) and on the same side as the object.
• Negative when the image is inverted (real) and on the opposite side of the lens compared to the object. Hence, for a convex lens,the value of magnification m is positive as well as negative.

the image is real and diminished formed by a convex lens

Reason — If the magnification produced by a lens is -0.5, it means the image is inverted compared to the object. The negative sign indicates inversion. The magnitude of 0.5 indicates that the image is smaller than the object.

less than 1

Reason — As the image formed by a concave lens is always smaller than the object, the magnitude of the magnification is always less than 1, indicating the diminished size of the image compared to the object.

positive, negative

Reason — If a lens deviates a ray towards its centre, its power is positive and if it deviates the ray away from its centre, its power is negative. Hence, power of convex lens is positive and that of concave lens is negative.

increases

Reason — As we know that,

$P = \dfrac{1}{f}$

Where,

P = power of lens and

f = focal length of lens.

Hence, power of a lens is inversely proportional to its focal length.

So, on reducing the focal length of a lens, its power increases.

convex of focal length 1.0 m

Reason — As we know that,

$P = \dfrac{1}{f}$

Where,

P = power of lens and

f = focal length of lens.

Given,

power = + 1.0 D

Substituting the values in the formula we get,

$+1.0 \text{ D} = \Big(\dfrac{1}{f}\Big)\text{ m} \\[0.5em] \Rightarrow f = 1.0 \text{ m}$

Hence, focal length = 1.0 m

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

Reason — Linear magnification m is given by m = $\dfrac{v}{u}$ where v = image distance and u = object distance.

decreases

Reason — A lens bends light less in water than in air because water and glass have similar refractive indices. This makes the focal length longer and the power of the lens decreases.

(a) As we know,

The formula for magnification of a lens is —

$m = \dfrac{\text{length of image (I)}}{\text{length of object (O)}} = \dfrac{v}{u} \\[0.5em]$

where,

v = image distance

u = object distance.

Given,

Height of a candle, O = 3 cm

Height of the image of candle, I = 6 cm

Image distance, v = 30 cm

Substituting the values in the formula we get,

$\dfrac{6}{3} = \dfrac{30}{u} \\[0.5em] u = \dfrac{30 \times 3}{6} \\[0.5em] u = 5 \times 3 \\[0.5em] \Rightarrow u = 15$

Therefore, object distance is equal to 15 cm.

(b) As we know, lens formula is —

$\dfrac{1}{v} – \dfrac{1}{u} = \dfrac{1}{f} \\[0.5em]$

Substituting the values in the formula we get,

$\dfrac{1}{30} – \dfrac{1}{-15} = \dfrac{1}{f} \\[0.5em] \dfrac{1}{30} + \dfrac{1}{15} = \dfrac{1}{f} \\[0.5em] \dfrac{2 + 1}{30} = \dfrac{1}{f} \\[0.5em] \dfrac{3}{30} = \dfrac{1}{f} \\[0.5em] f = \dfrac{30}{3} \\[0.5em] \Rightarrow f = 10$

Hence, the focal length of the lens, f = 10 cm.

(a) As the lens used is a concave lens, so the image formed will be erect and diminished. It will be virtual in nature as the image is formed on the same side as the object.

(b) As we know, the lens formula is —

$\dfrac{1}{v} – \dfrac{1}{u} = \dfrac{1}{f} \\[0.5em]$

Given

u = -20 cm

v = – 10 cm

Substituting the values in the formula we get,

$\dfrac{1}{-10} – \dfrac{1}{-20} = \dfrac{1}{f} \\[0.5em] -\dfrac{1}{10} + \dfrac{1}{20} = \dfrac{1}{f} \\[0.5em] \dfrac{-2 + 1}{20} = \dfrac{1}{f} \\[0.5em] -\dfrac{1}{20} = \dfrac{1}{f} \\[0.5em] \Rightarrow f = -20 cm$

Therefore, focal length of the lens = 20 cm (negative)

As we know,

The formula for magnification of a lens is —

$m = \dfrac{v}{u} \\[0.5em]$

Given,

f = + 25 cm

As image is virtual and magnified, so m = + 2

Substituting the values in the formula we get,

$+2 = \dfrac{v}{u} \\[0.5em] \Rightarrow v = 2u \\[0.5em]$

Therefore, we get, v = 2u

Now as we know, the lens formula is —

$\dfrac{1}{v} – \dfrac{1}{u} = \dfrac{1}{f} \\[0.5em]$

Substituting the values in the formula, we get,

$\dfrac{1}{2u} – \dfrac{1}{u} = \dfrac{1}{25} \\[0.5em] \dfrac{1-2}{2u} = \dfrac{1}{25} \\[0.5em] -\dfrac{1}{2u} = \dfrac{1}{25} \\[0.5em] 2u = - 25 \\[0.5em] \Rightarrow u = - 12.5 cm \\[0.5em]$

Therefore, object should be placed at a distance of 12.5 cm infront of the lens.

(i) As we know,

The formula for magnification of a lens is —

$m = \dfrac{v}{u} \\[0.5em]$

Given,

f = +0.12 m

Image is real and magnified, m = -3

Substituting the values in the formula we get,

$-3 = \dfrac{v}{u} \\[0.5em] \Rightarrow v = -3u \\[0.5em]$

Therefore, we get, v = -3u

Now as we know, the lens formula is —

$\dfrac{1}{v} – \dfrac{1}{u} = \dfrac{1}{f} \\[0.5em]$

Substituting the values in the formula, we get,

$\dfrac{1}{-3u} – \dfrac{1}{u} = \dfrac{1}{0.12} \\[0.5em] \dfrac{-1-3u}{3u} = \dfrac{1}{0.12} \\[0.5em] -\dfrac{4}{3u} = \dfrac{1}{0.12}\\[0.5em] u = - \dfrac{4 \times 0.12}{3} \\[0.5em] \Rightarrow u = - 0.16 m \\[0.5em]$

Therefore, object should be placed at a distance of 0.16 m infront of the lens.

As we know, the lens formula is —

$\dfrac{1}{v} – \dfrac{1}{u} = \dfrac{1}{f} \\[0.5em]$

Given,

Object lies at a distance 1.0 m from a screen.

v = 75 cm

So, u = - 25 cm

Substituting the values in the formula, we get,

$\dfrac{1}{75} – \dfrac{1}{-25} = \dfrac{1}{f} \\[0.5em] \dfrac{1}{75} + \dfrac{1}{25} = \dfrac{1}{f} \\[0.5em] \dfrac{1+3}{75} = \dfrac{1}{f} \\[0.5em] \dfrac{4}{75} = \dfrac{1}{f}\\[0.5em] \Rightarrow f = \dfrac{75}{4} \\[0.5em] \Rightarrow f = \text{18.75 cm} \\[0.5em]$

Therefore, focal length of the lens is 18.75cm.

ii) As we know,

the formula for magnification of a lens is —

$m = \dfrac{v}{u} \\[0.5em]$

Given,

v = 75 cm

u = - 25 cm

Substituting the values in the formula, we get,

$m = \dfrac{75}{-25} \\[0.5em] m = -3 \\[0.5em]$

Therefore, the magnification is -3.

(i) As we know, the lens formula is —

$\dfrac{1}{v} – \dfrac{1}{u} = \dfrac{1}{f} \\[0.5em]$

Given,

v = -60 cm

So, u = -15 cm

Substituting the values in the formula, we get,

$\dfrac{1}{-60} – \dfrac{1}{-15} = \dfrac{1}{f} \\[0.5em] - \dfrac{1}{60} + \dfrac{1}{15} = \dfrac{1}{f} \\[0.5em] \dfrac{1}{15} - \dfrac{1}{60} = \dfrac{1}{f} \\[0.5em] \dfrac{4-1}{60} = \dfrac{1}{f} \\[0.5em] \dfrac{3}{60} = \dfrac{1}{f}\\[0.5em] \Rightarrow f = \dfrac{60}{3} \\[0.5em] \Rightarrow f = 20cm \\[0.5em]$

Therefore, focal length of the lens is 20 cm.

ii) As we know,

the formula for magnification of a lens is —

$m = \dfrac{v}{u} \\[0.5em]$ Given,

v = -60 cm

u = -15 cm

Substituting the values in the formula, we get,

$m = \dfrac{- 60}{- 15} \\[0.5em] m = 4 \\[0.5em]$

Therefore, the magnification is 4.

(iii) The nature of the image is erect, virtual and magnified.

(a) As the image is formed on the other side of the lens, so the image is real. Hence, the lens is convex.

(b) (i) As we know, the lens formula is —

$\dfrac{1}{v} – \dfrac{1}{u} = \dfrac{1}{f} \\[0.5em]$

Given,

u = – 45 cm

v = + 90 cm

Substituting the values in the formula, we get,

$\dfrac{1}{90} – \dfrac{1}{-45} = \dfrac{1}{f} \\[0.5em] \dfrac{1}{90} + \dfrac{1}{45} = \dfrac{1}{f} \\[0.5em] \dfrac{1+2}{90} = \dfrac{1}{f} \\[0.5em] \dfrac{3}{90} = \dfrac{1}{f} \\[0.5em] \dfrac{1}{30} = \dfrac{1}{f}\\[0.5em] \Rightarrow f = 30cm \\[0.5em]$

Therefore, focal length of the lens is 30 cm.

ii) As we know,

the formula for magnification of a lens is —

$m = \dfrac{v}{u} \\[0.5em]$ Given,

u = –45 cm

v = +90 cm

Substituting the values in the formula, we get,

$m = \dfrac{90}{-45} \\[0.5em] m = -2 \\[0.5em]$

Therefore, the magnification is -2.

An inverted, real and same size image of an object, is formed by a convex lens when the object is placed at 2f i.e. u = 2f₁.

(a) When the object is at 2f₁ in a convex lens then the image is formed at 2f₂. Hence, position of the image is 60 cm behind the lens.

(b) To find the focal length of this lens, we use the relationship given below,

Object distance (u) = 2f

Given,

Object distance (u) = 60 cm

Using the two equivalent values we get,

$60 = 2 \times f \\[0.5em] f = \dfrac{60}{2} \\[0.5em] \Rightarrow f = 30 \\[0.5em]$

Therefore, the focal length of the given lens is 30 cm.

(a) As we know,

the formula for magnification of a lens is —

$m = \dfrac{v}{u} \qquad \bold{\text{(Equation 1)}}$

Given,

u = -30 cm

and

$m = \dfrac{I}{O} = \dfrac{1}{3} \qquad \bold{\text{(Equation 2)}}$

Substituting the values of u and Equation 2 in Equation 1, we get,

$\dfrac{v}{-30} = \dfrac{1}{3} \\[0.5em] v = -10 \\[0.5em]$

Therefore, the image is formed at 10 cm infront of the lens.

(b) As we know, the lens formula is —

$\dfrac{1}{v} – \dfrac{1}{u} = \dfrac{1}{f} \\[0.5em]$

Given,

u = -30 cm

v = -10 cm

The lens used is concave in nature.

Substituting the values in the formula, we get,

$\dfrac{1}{-10} – \dfrac{1}{-30} = \dfrac{1}{f} \\[0.5em] -\dfrac{1}{10} + \dfrac{1}{30} = - \dfrac{1}{f} \\[0.5em] \dfrac{-3+1}{30} = - \dfrac{1}{f} \\[0.5em] -\dfrac{2}{30} = \dfrac{1}{f} \\[0.5em] -\dfrac{1}{15} = \dfrac{1}{f} \\[0.5em] \Rightarrow f = -15 cm \\[0.5em]$

Therefore, the focal length is 15 cm (negative)

As we know,

$\text{Power} = \dfrac{1}{\text{focal length}}$

Given,

Power of the lens = +2.0D

As the given power is positive,

Therefore, we can say that the lens used is convex in nature.

Substituting the value of power in formula we get,

$\text{2.0} = \dfrac{1}{\text{focal length}} \\[0.5em] \Rightarrow \text{focal length} = \dfrac{1}{2} \\[0.5em] \Rightarrow \text{focal length} = \text{0.5m} \\[0.5em] \Rightarrow \text{focal length} = \text{50cm} \\[0.5em]$

Therefore, the focal length is 50 cm and the lens used is convex in nature.

As we know,

$\text{Power} = \dfrac{1}{\text{focal length}}$

Given,

focal length = 20 cm = 0.2m

Substituting the value of focal length in formula we get,

$\text{Power} = \dfrac{1}{\text{0.2 m}} \\[0.5em] \Rightarrow \text{Power} = 5 D \\[0.5em]$

As the lens used is concave in nature so power is negative.

Hence, Power = – 5D

As we know,

$\text{Power} = \dfrac{1}{\text{focal length}}$

Given,

focal length = 25 cm = 0.25m

Substituting the value of focal length in formula we get,

$\text{Power} = \dfrac{1}{\text{0.25 m}} \\[0.5em] \Rightarrow \text{Power} = 4 D \\[0.5em]$

As the lens used is convex in nature so power is positive.

Hence, P = +4D

As we know,

$\text{Power} = \dfrac{1}{\text{focal length}}$

Given,

Power of the lens = -2.0 D

As the given power is negative hence, we can say that the lens used is concave in nature.

Substituting the value of power in formula we get,

$-\text{2.0} = \dfrac{1}{\text{focal length}} \\[0.5em] \Rightarrow \text{focal length} = -\dfrac{1}{2} \\[0.5em] \Rightarrow \text{focal length} = -\text{0.5m} \\[0.5em] \Rightarrow \text{focal length} = -\text{50cm} \\[0.5em]$

Therefore, the focal length is 50 cm and the lens used is concave in nature.

The magnification (m) of the lens is given as -3, which implies that the image is real and inverted because m is negative for a real and inverted image.

The image is enlarged as the magnitude of magnification is greater than 1.

As the image formed is real, inverted and magnified so the lens used is convex in nature.

As we know, the relation between u and v is as shown below —

$m = \dfrac{v}{u} \\[0.5em]$

Given,

m = -3

Substituting the values in the formula

$-3 = \dfrac{v}{u} \\[0.5em] \Rightarrow v = -3u \\[0.5em]$

Hence, the lens is convex in nature and the image distance v = -3u.

As the magnification (m) of the lens is given as +0.5, which implies that the image is virtual and erect because m is positive for a virtual and erect image.

The image is smaller in size as the magnitude of magnification is smaller than 1.

As the image formed is virtual, erect and smaller in size so the lens used is concave in nature.

As we know, the relation between u and v is as shown below —

$m = \dfrac{v}{u} \\[0.5em]$

Given,

magnification by a lens is +0.5

Substituting the values in the formula

$0.5 = \dfrac{v}{u} \\[0.5em] \Rightarrow v = 0.5u \\[0.5em]$

The lens is concave in nature and the image distance v = 0.5u.

Given,

Focal length = f = 0.2m

So,

$\text{P} = \dfrac{1}{\text{focal length (in metres)}} \\[1em] = \dfrac{1}{0.2} \\[1em] = \dfrac{10}{2} \\[1em] = 5 \space \text{D}$

A lens with higher power has a shorter focal length, allowing it to focus on nearby objects more effectively. This makes it ideal for close-up shots.

As we know, the lens formula is —

$\dfrac{1}{v} – \dfrac{1}{u} = \dfrac{1}{f} \\[0.5em]$

Given,

f = - 30 cm

u = - 30 cm

Substituting the values in the formula, we get,

$\dfrac{1}{v} – \dfrac{1}{-30} = \dfrac{1}{-30} \\[0.5em] \dfrac{1}{v} + \dfrac{1}{30} = - \dfrac{1}{30} \\[0.5em] \dfrac{1}{v} = -\dfrac{1}{30} - \dfrac{1}{30} \\[0.5em] \dfrac{1}{v} = \dfrac{-1-1}{30} \\[0.5em] \dfrac{1}{v} = -\dfrac{2}{30} \\[0.5em] \dfrac{1}{v} = -\dfrac{1}{15} \\[0.5em] \Rightarrow v = -15$

Therefore, the image in formed 15 cm in front of the lens.

ii) As we know,

the formula for magnification of a lens is —

$m = \dfrac{v}{u} \\[0.5em]$ Given,

u = - 30 cm

v = - 15 cm

Substituting the values in the formula, we get,

$m = \dfrac{-15}{-30} \\[0.5em] m = 0.5 \\[0.5em]$

Therefore, the magnification is +0.5

The image is virtual as the magnification is positive.

i) As the image is formed on the other side of the lens, so the image is real. Hence, the lens is convex.

As we know, the lens formula is —

$\dfrac{1}{v} – \dfrac{1}{u} = \dfrac{1}{f} \\[0.5em]$

Given,

f = 10 cm

u = - 8 cm

Substituting the values in the formula, we get,

$\dfrac{1}{v} – \dfrac{1}{-8} = \dfrac{1}{10} \\[0.5em] \dfrac{1}{v} + \dfrac{1}{8} = \dfrac{1}{10} \\[0.5em] \dfrac{1}{v} = \dfrac{1}{10} - \dfrac{1}{8} \\[0.5em] \dfrac{1}{v} = \dfrac{4-5}{40} \\[0.5em] \dfrac{1}{v} = -\dfrac{1}{40} \\[0.5em] \Rightarrow v = - 40 cm \\[0.5em]$

Therefore, the image is formed 40 cm in front of the lens.

ii) As we know,

the formula for magnification of a lens is —

$m = \dfrac{v}{u} \\[0.5em]$ Given,

u = –8 cm

v = -40 cm

Substituting the values in the formula, we get,

$m = \dfrac{-40}{-8} \\[0.5em] m = 5 \\[0.5em]$

Therefore, the magnification is +5.0

As the magnification is positive so the image is erect.

We follow the cartesian sign convention to measure the distance in a lens according to which:

1. The optical centre of the lens is chosen as the origin of the coordinate system.
2. The object is considered to be placed on the principle axis to the left of the lens.
3. All the distances are measured along the principle axis from the optical centre of the lens. The distance of the object from the lens is denoted by u, the distance of the image by v and the distance of the second focus by f.
4. The distances measured in the direction of the incident ray are taken positive, while the distances opposite to the direction of incident ray are taken negative.
5. The length above the principle axis is taken positive, while the length below the principle axis is taken negative.
6. By sign convention, the focal length of the convex lens is positive and that of the concave lens is negative.
7. The distance of object (u) infront of lens is always negative. The distance of image (v) is positive if it is real and formed behind the lens, while it is negative if the image is virtual and formed in front of the lens.

Lens formula,

$\dfrac{1}{v} – \dfrac{1}{u} = \dfrac{1}{f}$

where, 'u' is the distance of the object from the optical centre of the lens, 'v' is the distance of the image from the optical centre while 'f' is the focal length, the distance between the optical centre and the focus of the lens.

The ratio of length of image I perpendicular to the principal axis, to the length of the object O, is called linear magnification.

The expression is:

$m = \dfrac{\text{length of image (I)}}{\text{length of object (O)}} = \dfrac{v}{u} \\[0.5em]$

where,

v = image distance

u = object distance.

The deviation of the incident light rays produced by a lens on refraction through it is a measure of its power.

The S.I. unit of power is dioptre (D).

Given, P₁ = +2.5D and

P₂ = -2.5D

We know P = P₁ + P₂

Substituting we get, P = +2.5 + (-2.5) = 0

Hence, the power of the combination of these two lenses is 0 D. This means that the combination behaves like a piece of plain glass. When light passes through this combination, it will not converge or diverge; it will simply pass straight through without being affected by refraction.

(i) When the focal length of a lens is positive, then the lens is a convex lens.

(ii) When the focal length of a lens is negative, then the lens is a concave lens.

(i) The positive sign of magnification indicates a virtual image while a negative sign indicates a real image.

(ii) The positive sign of magnification indicate an erect image while a negative sign indicates an inverted image.

The relation between power of a lens and its focal length is as follows —

$P = \dfrac{1}{f}$

Where,

P = power of lens and

f = focal length of lens.

Hence, power of a lens is inversely proportional to its focal length.

As we know that,

$P = \dfrac{1}{f}$

Where,

P = power of lens and

f = focal length of lens.

Hence, power of a lens is inversely proportional to its focal length.

$P = \dfrac{1}{2f}$

So, if the focal length is doubled, the power gets halved.

Depending on the direction in which a lens deviates the light rays, its power is either positive or negative.

If a lens deviates a ray towards its centre (converges), the power is positive and if it deviates the ray away from its centre (diverges), the power is negative.

Hence, the power of a convex lens is positive and of a concave lens is negative.

It is a concave lens

When comparison is done between a thick lens and a thin lens then, a thick lens has more power than a thin lens.

A thick lens (i.e. a lens having surfaces of more curvature) is of short focal length and it deviates the rays more, while a thin lens (i.e. a lens having surfaces of less curvature) is of large focal length and deviates the rays less as power of a lens is reciprocal of its focal length.

Below is the labelled ray diagram to show the formation of an image by a magnifying glass:

Three characteristics of the formed image are:

1. Image is virtual
2. Image is erect
3. Image is magnified

The principle used for determination of approximate focal length of a convex lens is as follows —

A beam of parallel rays from a distant object incident on a convex lens gets converged in the focal plane of the lens.

In an open space, against a white wall, place a metre rule horizontally with its 0 cm end touching the wall, with its other end towards the illuminated object at a very large distance.

By moving the convex lens to and fro along the length of the metre rule, focus the object on the wall. Since, the light rays incident from a distant object are nearly parallel, the image of it formed on the wall is almost at the focus of the lens.

So, from the image, the distance of the lens is read directly from the metre scale. This gives the approximate focal length of the lens.

(i) Below ray diagram shows the formation of image I at O itself:

(ii) The size of the image is same as that of the object O.

(iii) The image I is inverted and real.

(iv) The distance of object O from the optical centre of the lens will be equal to the focal length of the lens.

(v) The position of plane mirror relative to the lens does not affect the position of image as long as rays fall normally on the plane mirror M.

The equipments required to determine the focal length by using a plane mirror are —

1. A vertical stand
2. A plane mirror
3. A lens
4. A pin

(i) We place the lens L horizontally on a plane mirror MM’. Set the pin P in the clamp, in such a way that the tip of the pin is vertically above the centre O of the lens and is perfectly horizontal.

(ii) Set the height of the pin till it has no parallax (i.e., if the pin and its image shift together) so that its inverted image as seen, vertically above the pin.

(ii) Now we measure the distance x of the pin P from the lens and the distance y of the pin from the mirror, using a metre rule and a plumb line.

(iv) Now we calculate the average of the two distances. This gives the focal length of the lens, i.e.,

$f = \dfrac{(x + y)}{2}$

25 cm

Reason — The least distance of distinct vision of normal eye is 25 cm.

a virtual and magnified image

Reason — A magnifying glass, which is typically a convex lens, is used to produce a magnified, virtual image.

6

Reason — As we know,

$m = 1 + \dfrac{D}{f}$

Given,

f = +5 cm

D is the distance of the distinct vision, D = 25 cm

Substituting the values in the formula we get,

$m = 1 + \dfrac{25}{5} \\[0.5em] m = 1 + 5 \\[0.5em] m = 6 \\[0.5em]$

Hence, the maximum magnifying power of a convex lens of focal length 5 cm is 6.

less than 1'

Reason — The human eye is not able to see an object distinctly if it subtends an angle less than 1' = $\dfrac{1}{60°}$ at it.

convex lens of short focal length

Reason — A simple microscope, also known as a magnifying glass, typically consists of a single convex lens of short focal length because it provides greater magnification.

M = 1 + $\dfrac{D}{f}$

Reason — Magnifying power of a microscope is given as M = 1 + $\dfrac{D}{f}$.

convex, concave

Reason — Long-sighted individuals require convex lenses to converge light rays and focus them properly on the retina for clear vision of nearby objects, while short-sighted individuals require concave lenses to diverge light rays and focus them properly on the retina for clear vision of distant objects.

A ⟶ convex, B ⟶ concave

Reason — Convex lenses are used for magnifying objects held close to them, hence, lens A will be convex lens. Concave lenses produce upright and virtual images for objects placed beyond their focal point, hence, B is a concave lens.

A magnifying glass is a convex lens of short focal length fitted in a steel (or plastic) frame provided with a handle.

Human eye has a limitation that in order to see an object by the naked eye, it is necessary to place it at least at a distance of 25 cm from the eye.

So, we use a magnifying glass, infront of our eyes so as to get an erect virtual and magnified image of the object.

Two uses of a magnifying glass are as follows —

1. It is used by watchmakers, to see the small parts and screws of the watch.
2. It is used to see and read the small alphabets or figures.

In reference to the principal focus, the object is placed between the lens and the principal focus in order to obtain an enlarged image.

The image is obtained at the least distance of distinct vision D. It is formed between the lens and the principal focus.

The expression for the magnifying power (M) of a simple microscope is as follows —

$M = 1 + \dfrac{D}{f}$

where f is the focal length of the lens, and

D is the least distance of distinct vision.

The magnifying power of the microscope can be increased by using the lens of short focal length (i.e., shorter the focal length, more is the magnifying power). But it cannot be increased indefinitely.

Applications of a convex lens are —

(i) The objective lens of a telescope, camera, slide projector, etc., is a convex lens which forms the real and inverted image of the object.

(ii) Our eye lens is also a convex lens. The eye lens forms the inverted image of the object on retina.

Applications of concave lens are —

(i) A concave lens is used as the eye lens in a Galilean telescope to obtain the final erect image of the object.

(ii) A person suffering from short sightedness or myopia (i.e., unable to see the far objects distinctly) wears spectacles having the concave lens.

(i) On seeing a distant object through the lens, if its inverted image is seen, the lens is convex and if the upright image is seen, the lens is concave.

(ii) On keeping the lens near a printed page, if letters appear magnified, the lens is convex and if the letters appear diminished, the lens is concave.

When the colour of incident light changes from violet to red, the focal length of the lens increases. This is because violet light bends more than red light when passing through a lens.
As a result, violet rays focus closer to the lens, while red rays focus farther away. So, the lens has different focal lengths for different colours of light.
This defect is called chromatic aberration.

(Chromatic aberration is the inability of a lens to bring all colours of light to the same focus point, causing coloured fringes around images.)