Reflection at Curved Mirror

SYLLABUS

Real and Virtual images, Mirror formula

What is This Chapter About?

Stand in front of a concave shaving mirror and your face appears magnified. Look into the back of a shiny spoon and you see yourself upside down. Glance at the convex mirror mounted at a blind corner of a road and you see a wide-angle view of traffic. All of these experiences involve reflection at curved mirrors — surfaces that curve inward (concave) or outward (convex) and produce images with properties very different from flat plane mirrors.

Chapter 14 of NEB Grade 11 Physics introduces the quantitative study of image formation by curved mirrors. It is one of the most practically relevant chapters in geometrical optics — curved mirrors are used in telescopes, car headlights, satellite dishes, security systems, dental examinations, and countless other applications. Carrying 2 teaching hours as a concise but important chapter, it establishes the mirror formula and magnification relationships that are essential tools for solving optical problems.

Key Concepts You Will Learn

Types of Curved Mirrors

A concave mirror curves inward like the inside of a bowl. It converges parallel rays of light to a real focal point in front of the mirror. A convex mirror curves outward like the outside of a ball. It diverges parallel rays of light as if they came from a virtual focal point behind the mirror.

The principal axis is the line passing through the centre of curvature C and the pole P of the mirror. The focal point F lies on the principal axis at a distance equal to half the radius of curvature: [latex]f = \frac{R}{2}[/latex]. This relationship — that the focal length is half the radius of curvature — is derived from geometry and is fundamental to all curved mirror calculations.

Real and Virtual Images

A real image is one formed by actual convergence of reflected rays — it can be projected onto a screen and is always inverted. Real images are formed by concave mirrors when the object is beyond the focal point.

A virtual image is one from which reflected rays appear to diverge — it cannot be projected onto a screen and is always upright. Virtual images are formed by convex mirrors for all object positions, and by concave mirrors when the object is between the focal point and the mirror (inside the focal length).

Mirror Formula

The mirror formula relates the object distance u, image distance v, and focal length f of a curved mirror:

[latex]\frac{1}{v} + \frac{1}{u} = \frac{1}{f}[/latex]

This compact equation, derived from the geometry of ray reflection at a curved surface, applies to both concave and convex mirrors provided the standard sign convention is followed consistently. The New Cartesian Sign Convention takes distances measured in the direction of incident light as positive and distances measured against incident light as negative, with the pole of the mirror as the origin.

Magnification

The linear magnification produced by a curved mirror is defined as the ratio of image height to object height:

[latex]m = \frac{h_i}{h_o} = -\frac{v}{u}[/latex]

The negative sign in the magnification formula follows from the sign convention. If m is positive, the image is upright (virtual). If m is negative, the image is inverted (real). If |m| > 1, the image is magnified. If |m| < 1, the image is diminished.

Image Formation by Concave Mirrors

The nature, position, and size of the image formed by a concave mirror depend critically on where the object is placed relative to the focal point and centre of curvature. Key cases include: object at infinity (image at F, real, point-sized), object beyond C (image between F and C, real, inverted, diminished), object at C (image at C, real, inverted, same size), object between C and F (image beyond C, real, inverted, magnified), object at F (image at infinity — reflected rays are parallel), and object between F and P (image behind mirror, virtual, upright, magnified).

Understanding these cases and being able to draw accurate ray diagrams for each is a core skill developed in this chapter.

Applications of Curved Mirrors

Concave mirrors are used as shaving and makeup mirrors (object inside focal length — virtual, upright, magnified image), in torch lights and car headlights (source at focal point — reflected rays emerge as a parallel beam), in reflecting telescopes (collecting and focusing parallel rays from distant stars), and in solar furnaces (concentrating sunlight to very high temperatures at the focal point).

Convex mirrors are used as rear-view mirrors in vehicles (always produce upright, diminished, virtual images regardless of object distance — providing a wide field of view), and as security mirrors in shops and at road junctions.

Why This Chapter Matters for NEB Exams

Reflection at Curved Mirrors is examined regularly in NEB Grade 11 Physics. The mirror formula and magnification formula — applying them to find image position, nature, and size — generate standard numerical problems. Drawing accurate ray diagrams for different object positions is a classic examination question. The derivation of [latex]f = \frac{R}{2}[/latex] and the mirror formula itself appear as short derivation questions. Applications of concave and convex mirrors are popular short-answer topics.

Important Topics to Focus On

• Types of curved mirrors and their properties — concave and convex
• Key terms: pole, principal axis, centre of curvature, focal point, focal length
• Relation between focal length and radius of curvature: [latex]f = \frac{R}{2}[/latex]
• Distinction between real and virtual images
• New Cartesian Sign Convention for curved mirrors
• Mirror formula: [latex]\frac{1}{v} + \frac{1}{u} = \frac{1}{f}[/latex] — derivation and application
• Magnification: [latex]m = -\frac{v}{u}[/latex]
• Image formation by concave mirror for all object positions — ray diagrams
• Image formation by convex mirror — always virtual, upright, diminished
• Applications of concave mirrors (torches, telescopes, solar furnaces)
• Applications of convex mirrors (rear-view mirrors, security mirrors)
• Numerical problems on mirror formula and magnification

Connection to Other Chapters

Reflection at Curved Mirrors connects to Chapter 15 (Refraction at Plane Surfaces) and Chapter 17 (Lenses) — together these three chapters form the core of geometrical optics, using the same sign conventions and analogous formulae. The mirror formula is structurally identical to the lens formula, making the two chapters strongly complementary. In Grade 12, curved mirror principles connect to Chapter 10 (Nature and Propagation of Light) — where Huygens’ Principle explains reflection — and to Chapter 12 (Diffraction) — where the resolving power of reflecting telescopes is discussed. The parabolic reflectors used in satellite dishes and radio telescopes are curved mirror applications on an enormous scale.

Optics

The study of nature and propagation of light is called optics. This branch of physics further classifies into two types: (i) Ray optics and (ii) Wave optics.

Ray optics deals with particle nature of light whereas wave optics considers light as a wave.

Ray Optics

The study of particle nature of light comes under ray optics. It works under following assumptions:

(i) Rectilinear propagation of light

(ii) Laws of reflection

(iii) Laws of refraction

Laws of Reflection:

1. The incident rays, the reflected rays and the normal at the point of incidence, all the three lies in one plane.
2. The angle of incidence and the angle of reflection are equal to one another.

i.e., for a normal incidence, ∠i = 0, ∠r = 0. Hence, a ray of light falling normally get back to its original path.

Spherical Mirrors:

A spherical mirror, is a part of a hollow sphere with one of its sides highly silvered.

If the outer surface of the part of the sphere is silvered and inner surface becomes reflecting, such mirror is called concave mirror. If the inner surface of the part of the sphere is silvered and outer surface becomes reflecting, such mirror is called the concave mirror.

Some Definitions

1. Centre of Curvature (C): The centre of curvature of a spherical mirror is the centre of the sphere of which the mirror is a part.
2. Radius of Curvature (R): The radius of curvature of a spherical mirror is the radius of the sphere of which the mirror is a part.
3. Pole of the mirror (P): The centre (or the middle point) of the spherical mirror is called the pole.
4. Principal axis: The line passing through the centre of curvature (C) and the pole (P) of the spherical mirror is called the principal axis.
5. Focus (F): The focus of the spherical mirror is the point on the principal axis at which the incident rays parallel to the principal axis actually meet or appears to meet after reflection. For concave mirror, the incident rays parallel to the principal axis actually meet at the focus & in case of convex mirror, the incident rays parallel to the principal axis don’t actually meet but they appear to be coming from the focus after reflection.

6. Focal length (f): The focal length (f) of a spherical mirror is the distance between the focus (F) and the pole (P) of the mirror. In the figure above, the distance, PF = f is the focal length.

The Concave Mirror:

Sign convention:

1. All distances are measured from the pole of the mirror.
2. Positive sign is used for real distances.
3. Negative sign is used for virtual distances.
4. For real image, magnification is positive & for virtual image, it is negative.
5. The focal length & the radius of curvature are always positive.

Relation between R and f:

Let us consider, a concave mirror having focal length f and radius of curvature (R). Let, a ray of light AB parallel to the principal axis incident at B & after reflection, it passes through the focus (F). Let, C & P be the centre of curvature & the pole of the mirror.

In the figure above, ∠i = ∠ABC = angle of incidence.

∠r = ∠CBF = angle of reflection

And, from the laws of reflection,

∠i = ∠r

i.e. ∠ABC = ∠CBF …………………. (1)

But, ∠ABC = ∠BCF [∵alternate angle]

[latex]\therefore[/latex] ∠BCF = ∠CBF

So, [latex]\Delta[/latex]CBF is an isosceles triangle.

[latex]\therefore[/latex] CF = FB

If the aperture of the mirror is small, then, point B approaches to P. Then,

FB = FP

So, PC = CF + FP

            = FB + FP

Or, R = f + f

[latex]\therefore[/latex] R = 2f. 

Hence, the focal length of the concave mirror is a half of its radius.

Differences between real and virtual image

Mirror Formula

Concave mirror

i. When real image is formed

Let us consider, a concave mirror of focal length f and radius of curvature R. Let, an object AB is placed in front of it. A ray of light AR parallel to the principal axis, after reflection, passes through focus (F). Another ray AC passing through centre of curvature retraces its path. These rays intersect at a point A’, where a real, inverted image A’B’ is formed.

From figure, PF = f = focal length

PB = u = object distance

PB’ = v = Image distance.

Δs ABC & A’B’C are similar triangle,

[latex]\dfrac{AB}{A’B’} = \dfrac{BC}{B’C}[/latex] ……….. (i)

Again, Δs RSF & A’B’F are similar triangles.

[latex]\dfrac{RS}{A’B’} = \dfrac{FS}{FB’}[/latex]

Since, RS = AB,

[latex]\dfrac{AB}{A’B’} = \dfrac{FS}{FB’}[/latex] ……………… (ii)

From equations (i) & (ii), we get,

[latex]\dfrac{BC}{B’C} = \dfrac{FS}{FB’}[/latex]

If aperture of the mirror is small, then, points S and P will be very close to each other. So,

[latex]FS \approx FP[/latex]

Then, above equation becomes,

[latex]\dfrac{BC}{B’C} = \dfrac{FP}{FB’}[/latex]

Or, [latex]\dfrac{PB-PC}{PC-PB’} = \dfrac{FP}{PB’-FP}[/latex]

Or, [latex]\dfrac{u-R}{R-v} = \dfrac{f}{v-f}[/latex]

Or, [latex]\dfrac{u-2f}{2f-v} = \dfrac{f}{v-f}[/latex]

Or, [latex]uv – uf – 2fv + 2f^2 = 2f^2 – fv[/latex]

Or, [latex]uv = -fv + uf + 2fv[/latex]

Or, [latex]uv = fv + uf[/latex]

Dividing both sides by, uvf, we get,

[latex]\dfrac{uv}{uvf} = \dfrac{fv}{uvf} + \dfrac{uf}{uvf}[/latex]

Or, [latex]\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}[/latex]

This is the required mirror formula for concave mirror as the image formed is real.

ii. When virtual image is formed

Let, MN be the small aperture of concave mirror. An object AB is held perpendicular to the principal axis of the mirror between the focus and pole (P) as shown in figure. A virtual image (erect & magnified) of the object is formed as shown in figure. From figure,

PC = R = radius of curvature

PF = f = focal length

PA = u = object distance

PA^‘ = -v = image distance

In [latex]\Delta ABC[/latex] ~ [latex]\Delta A’B’C'[/latex]

[latex]\frac{AB}{A’B’} = \frac{AC}{A’C}[/latex] …………………………………… (1)

Again, from [latex]\Delta ABP[/latex] ~ [latex]\Delta A’B’P[/latex],

[latex]\frac{AB}{A’B’} = \frac{PA}{PA’}[/latex] ………………………….. (2)

From eq^ns. (1) & (2), we get,

[latex]\frac{AC}{A’C} = \frac{PA}{PA’}[/latex]

Or, [latex]\frac{PC – AP}{PC + PA’} = \frac{PA}{PA’}[/latex]

Or, [latex]\frac{R – u}{R – v} = \frac{u}{-v}[/latex]

Or, -vR + uv = uR – uv

Or, vR + uR = 2uv

Dividing both sides by uvR, we get,

[latex]\frac{2uv}{uvR} = \frac{uR}{uvR} + \frac{vR}{uvR}[/latex]

Or, [latex]\frac{2}{R} = \frac{1}{v} + \frac{1}{u}[/latex]

Or, [latex]\frac{2}{2f} = \frac{1}{u} + \frac{1}{v}[/latex]

Or, [latex]\frac{1}{f} = \frac{1}{u} + \frac{1}{v}[/latex]

This is the required mirror formula for concave mirror as the image formed is virtual.

Image formed by a concave mirror

a. When the object is at infinity

Image is formed at focus when the object is at infinity. i.e. if u = ∞,

[latex]\frac{1}{f} = \frac{1}{u} + \frac{1}{v}[/latex]

[latex]\therefore \frac{1}{f} = \frac{1}{\infty} + \frac{1}{v}[/latex]

Or, v = f 

Nature of Image

1. Real
2. Highly diminished.

b. When the object is beyond C

Image is formed between focus (F) and centre of curvature (C), when the object is placed beyond C.

Nature of image

1. Real
2. Diminished
3. Inverted

c. When the object is at C

Image is formed at the centre of curvature (C) when the object is placed at centre of curvature (C). i.e., when u = R = 2f, then,

[latex]\frac{1}{f} = \frac{1}{u} + \frac{1}{v}[/latex]

[latex]\frac{1}{f} = \frac{1}{2f} + \frac{1}{v}[/latex]

Or, [latex]\frac{1}{v} = \frac{1}{f} – \frac{1}{2f}[/latex]

Or, v = 2f = R 

Nature of Image

1. Real
2. Inverted
3. Same size of the object.

d. Object between F and C

The image is formed beyond centre of curvature (C) when the object is placed between focus (F) and centre of curvature (C).

Nature of Image

1. Real
2. Inverted
3. Magnified

e. Object at F

The image is formed at infinity when the object is placed at focus (F).

Nature of Image

1. Real
2. Inverted
3. Highly magnified

We have, u = f, then,

[latex]\frac{1}{f} = \frac{1}{u} + \frac{1}{v}[/latex]

[latex]\therefore \frac{1}{f} = \frac{1}{f} + \frac{1}{v}[/latex]

Or, [latex]\frac{1}{v} = 0 [/latex]

Or, v = ∞

f. Object between F & P

The image is formed on the other side of the mirror, when the object is placed between focus (F) and pole (P). In the ray diagram, we can see that, the reflected rays do not meet at a point but it seems to appear from the opposite side of the mirror. So,

Nature of Image

1. Virtual
2. Erect
3. Magnified.

Convex Mirror

Sign Conventions

1. All distances are measured from the pole of the mirror.
2. Positive sign is used for real distances.
3. Negative sign is used for virtual distances.
4. Magnification is always taken as negative.
5. The focal length & the radius of curvature are always taken as negative.

Relation between R and f

Let us consider, a ray of light AB parallel to the principal axis is incident at B on convex lens of focal length f and radius of curvature R. After reflection, it passes through BD and it appears to be coming from the focus F.

Here, ∠i = ∠ABN = angle of incidence.

∠r = ∠DBN = angle of reflection.

Now, ∠CBF = ∠DBN = ∠r (VOA).

And, ∠ABN = ∠BCF = ∠i (Corresponding angle).

∵ ∠i = ∠r

So, CF = BF. Now, if the aperture of the mirror is small, then, B and P are very close. So, FP = FB = CF. Thus,

CP = CF + FP = PF + PF

[latex]\therefore[/latex] R = f + f.

[latex]\therefore[/latex] R = 2f.

Or, f = [latex]\frac{R}{2}[/latex]

i.e., focal length of convex mirror is half of the radius of curvature.

Mirror formula for convex mirror

Let, MN be the small aperture of mirror. A ray of light from an object AG parallel to principal axis incident at point G and reflected along GQ. Another ray of light AP is reflected along PD. Third ray AE is reflected back. When these rays are produced back, they are intersecting at a point A’, where, a virtual, erect and diminished image (A’B’) is formed.

From figure, PB = u = object distance.

PC = R = radius of curvature.

PB = -v = image distance for virtual image.

PF = f = focal length.

In [latex]\Delta ABC[/latex] ~ [latex]\Delta A’B’C[/latex],

[latex]\frac{AB}{A’B’} = \frac{BC}{B’C}[/latex] ………………… (i)

Again, in [latex]\Delta ABP[/latex] ~ [latex]\Delta A’B’P[/latex]

[latex]\frac{AB}{A’B’} = \frac{PB}{PB’}[/latex] ……………….. (ii)

From eq^n. (i) and (ii), we get,

[latex]\frac{BC}{B’C} = \frac{PB}{PB’}[/latex]

Or, [latex]\frac{PC+PB}{PC – PB’} = \frac{PB}{PB’}[/latex]               [∵ PB’ = -v, for virtual image]

Or, [latex]\frac{-R + u}{-R+v} = \frac{u}{-v}[/latex]

Or, vR – uv = -uR + uv

Or, vR + uR = 2uv

Dividing both sides by uvR, we get,

Or, [latex]\frac{vR}{uvR} + \frac{uR}{uvR} = \frac{2uv}{uvR}[/latex]

Or, [latex]\frac{1}{u} + \frac{1}{v} = \frac{2}{R}[/latex]

[latex]\frac{1}{u} + \frac{1}{v} = \frac{2}{2f}[/latex]

[latex]\therefore \frac{1}{u} + \frac{1}{v} = \frac{1}{f}[/latex].

This is the required mirror formula for convex mirror as the image formed is virtual.

Linear Magnification

It is defined as the ratio of size of the image obtained by mirror to the size of the object. It is denoted by m.

[latex]\therefore m = \frac{size\ of\ image\ (I)}{size\ of\ object\ (O)}[/latex]

It can also be defined as the ratio of image distance to the object distance.

Linear magnification (m) = [latex]\frac{v}{u}[/latex]

Magnification is positive for real image and negative for virtual image.

Application of spherical mirror

a. Concave Mirror

1. Reflecting telescope.
2. Dental mirror.
3. Make-up and shaving mirror.

b. Convex mirror

1. Driving mirror as rear view of vehicles.
2. Safe view at dangerous corners.

1. At what position an object be placed in front of a concave mirror of radius of curvature 0.4 m so that an erect image of magnification 3 be produced? At what position an object be placed in front of a concave mirror of radius of curvature 0.4 m so that an erect image of magnification 3 be produced?

Solution
Given,
Object distance (u) =?
Radius of curvature (R) = 0.4 m
Magnification (m) = 3
We know,
Focal length (f) = [latex]\frac{R}{2} = \frac{0.4}{2} = 0.2[/latex] m

Now, magnification (m) = [latex]\frac{v}{u}[/latex]

Or, 3 = [latex]\frac{v}{u}[/latex]

Or, v = 3u
Since, image is erect, v = -3u
Now, using mirror formula,
[latex]\frac{1}{f} = \frac{1}{u} + \frac{1}{v}[/latex]

Or, [latex]\frac{1}{0.2} = \frac{1}{u} – \frac{1}{3u}[/latex]

Or, [latex]\frac{1}{0.2} = \frac{3 – 1}{3u}[/latex]

Or, [latex]\frac{1}{0.2} = \frac{2}{3u}[/latex]

Or, 3u = 0.4
Or, u = [latex]\frac{0.4}{3}[/latex] 

= 0.133 m = 13.3 cm

2. A pole 4 m long is laid along the principal axis of a convex mirror of focal length 1 m. The end of the pole nearer the mirror is 2m from it. Find the length of image of the pole.

Solution
Given,
Length of pole (l) = 4m
Focal length of convex mirror (f) = -1 m
Nearer end:
Object distance (u) = 2m

Using mirror formula,
[latex]\frac{1}{f} = \frac{1}{u} + \frac{1}{v}[/latex]

Or, [latex]-\frac{1}{1} = \frac{1}{2} + \frac{1}{v_1}[/latex]

Or, [latex]\frac{1}{v_1} = -1 – \frac{1}{2} = – \frac{3}{2}[/latex]

Or, v₁ = [latex]-\frac{2}{3}[/latex] = – 0.667 m
Again, far end:
Object distance (u’) = 6 m
Using mirror formula,
[latex]\frac{1}{f} = \frac{1}{u} + \frac{1}{v_2}[/latex]

Or, [latex]-\frac{1}{1} = \frac{1}{6} + \frac{1}{v_2}[/latex]

Or, [latex]\frac{1}{v_2} = -1-\frac{1}{6} = -\frac{7}{6}[/latex]

Or, v₂ = [latex]-\frac{6}{7}[/latex] = – 0.857 m

Now, length of image of pole ([latex]\Delta v[/latex])  = v₂ – v₁
      = – 0.857 – ( – 0.667)
      = – 0.857 + 0.667
      = – 0.19 m.

3. A mirror forms an erect image 30 cm from the object and twice its height. What is its radius of curvature?

Solution

Given,

Distance between object and image (d) = 30 cm

Let, object distance (u) = x

So, image distance (v) = 30 – x

According to question,

Magnification (m) = 2

Or, [latex]\dfrac{v}{u} = 2[/latex]

Or, [latex]\dfrac{30-x}{x} = 2[/latex]

Or, 30 – x = 2x

Or, 30 = 3x

Or, x = 10 cm

i.e. object distance (u) = 10 cm

and, image distance (v) = 30 – x = 30 – 10 = 20 cm

But the image is erect, so, v = – 20 cm

Now, Using mirror formula,

[latex]\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}[/latex]

[latex]= \dfrac{1}{10} – \dfrac{1}{20}[/latex]

[latex]= \dfrac{1}{20}[/latex]

∴ f = 20 cm

So, radius of curvature (R) = 2f = 2 x 20 = 40 cm = 0.4 m.

4. An erect image, three times the size of the object is obtained with a concave mirror of radius of curvature 36 cm. What is the position of the object?

Solution

Given,

Magnification (m) = 3

Radius of curvature (R) = 36 cm

∴ f = [latex]\dfrac{R}{2} = \dfrac{36}{2} = 18[/latex] cm

Object distance (u) =?

We know,

m = [latex]\dfrac{v}{u}[/latex]

Or, 3 = [latex]\dfrac{v}{u}[/latex]

Or, v = 3u

Or, v = -3u [∵ image is erect]

Using mirror formula, we get,

[latex]\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}[/latex]

Or, [latex]\dfrac{1}{18} = \dfrac{1}{u} + \dfrac{1}{(-3u)}[/latex]

Or, [latex]\dfrac{1}{18} = \dfrac{1}{u} – \dfrac{1}{3u}[/latex]

Or, [latex]\dfrac{1}{18} = \dfrac{3-1}{3u}[/latex]

Or, [latex]\dfrac{1}{18} = \dfrac{2}{3u}[/latex]

Or, 3u = 36

Or, u = [latex]\dfrac{36}{3}[/latex] = 12 cm

5. An object 10 cm high is placed infront of a convex mirror of focal length 20 cm and the object is 30 cm from the mirror. Find the height of the image.

Solution

Given,

Height of object (O) = 10 cm

Focal length (f) = – 20 cm

Object distance (u) = 30 cm

Height of the image (I) =?

We know that,

[latex]\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}[/latex]

Or, [latex]-\dfrac{1}{20} = \dfrac{1}{30} + \dfrac{1}{v}[/latex]

Or, [latex]-\dfrac{1}{20} – \dfrac{1}{30} = \dfrac{1}{v}[/latex]

Or, [latex]-\dfrac{1}{12} = \dfrac{1}{v}[/latex]

Or, v = – 12 cm

Again, we know,

m = [latex]\dfrac{I}{O} = \dfrac{v}{u}[/latex]

Or, [latex]\dfrac{I}{10} = \dfrac{-12}{30}[/latex]

Or, I = – 4 cm.

Hence, the size of the image is 4 cm.

6. Calculate the focal length of a concave mirror when an object placed at a distance of 40 cm makes image equal to the size of the object.

Solution

Given,

Focal length (f) =?

Object distance (u) = 40 cm

Since, image and object size is same, magnification (m) = 1

We know that,

m = [latex]\dfrac{v}{u}[/latex]

Or, 1 = [latex]\dfrac{v}{40\ cm}[/latex]

Or, v = 40 cm

Again, we have,

[latex]\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}[/latex]

Or, [latex]\dfrac{1}{f} = \dfrac{1}{40} + \dfrac{1}{40}[/latex]

Or, [latex]\dfrac{1}{f} = \dfrac{2}{40}[/latex]

Or, [latex]\dfrac{1}{f} = \dfrac{1}{20}[/latex]

Or, f = 20 cm

Hence, the focal length of concave mirror is 20 cm.

7. A metre scale is placed along the axis of a convex mirror of focal length 25 cm, its nearer end being at a distance of 50 cm. Calculate the size of the image formed.

Solution

Given,

Length of pole (l) = 1m = 100 cm

Focal length of convex mirror (f) = – 25 cm

Nearer end:

Object distance (u) = 50 cm

Using mirror formula,

[latex]\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}[/latex]

Or, [latex]-\dfrac{1}{25} = \dfrac{1}{50} + \dfrac{1}{v_1}[/latex]

Or, [latex]\dfrac{1}{v_1} = -\dfrac{1}{25} – \dfrac{1}{50} = -\dfrac{3}{50}[/latex]

Or, [latex]v_1 = -\dfrac{50}{3} = -16.667\ cm[/latex]

Again, far end:

Object distance (u’) = 150 cm

Using mirror formula,

[latex]\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v_2}[/latex]

Or, [latex]-\dfrac{1}{25} = \dfrac{1}{150} + \dfrac{1}{v_2}[/latex]

Or, [latex]\dfrac{1}{v_2} = -\dfrac{1}{25} – \dfrac{1}{150} = -\dfrac{7}{150}[/latex]

Or, [latex]v_2 = -\dfrac{150}{7} = -21.429\ cm[/latex]

Now, length of image of pole (Δv) = v2 – v1

= – 21.429 – ( – 16.667)

= – 21.429 + 16.667

= – 4.762 cm.

8. A convex mirror with a radius of curvature 30 cm forms a real image 20 cm from its pole. Explain how it is possible and find whether the image is erect or inverted.

Solution

Radius of curvature (R) = – 30 cm

Image distance (v) = 20 cm

Focal length (f) = [latex]\dfrac{R}{2} = -\dfrac{30}{2} = -15\ cm[/latex]

We know that,

[latex]\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}[/latex]

Or, [latex]-\dfrac{1}{15} = \dfrac{1}{u} + \dfrac{1}{20}[/latex]

Or, [latex]-\dfrac{1}{15} – \dfrac{1}{20} = \dfrac{1}{u}[/latex]

Or, [latex]\dfrac{1}{u} = -\dfrac{7}{60}[/latex]

Or, u = [latex]-\dfrac{60}{7} = -8.571[/latex] cm

Hence, it is possible when the object is virtual and it is at a distance of 8.571 cm from the mirror.