Curved Mirror Formula & Refraction at Spherical Surface (With Examples)

26.3 Formulae for Curved Mirrors

26.3.1 Formula for Curved Mirrors

Figure 26.11 A ray diagram illustrating how an object placed in front of a concave mirror forms an image. The distances u, v, and focal length f are related by the mirror formula.

An object is placed at distance u in front of a concave mirror of focal length f. The image produced by the mirror is formed at distance v from the mirror, as shown in Figure 26.11. We can show that u, v and f are related to one another by the formula

1/f = 1/u + 1/v

Derivation:

Figure 26.12 Ray diagram showing a point object O, the centre of curvature C, and the formation of an image in a concave mirror.

A point object O is placed at distance u from a concave mirror and on the principal axis, as shown in Figure 26.12. C is the centre of curvature of the mirror.

A ray from O is incident on the mirror at point A. The ray is reflected and a point image I is formed on the principal axis at distance v from the mirror. The reflected ray passes through I.

Using ΔAOI, we get:

φ = α + 2β      (1)

Using ΔACI, we get:

φ = γ + β

2φ = 2γ + 2β      (2)

(2) − (1),

φ = 2γ − α

α + φ = 2γ

tan α = AQ / OQ

tan φ = AQ / IQ

tan γ = AQ / CQ

If the point A is quite close to the centre P of the mirror, then the angles α, φ and γ are very small. Then we can make approximation.

Note:

1/f = 1/u + 1/v

f = focal length
u = object distance
v = image distance

Derivation (continued)

tan α ≈ α = AQ / OQ
tan φ ≈ φ = AQ / IQ
tan γ ≈ γ = AQ / CQ

α + φ = 2γ

We have

AQ/OQ + AQ/IQ = 2(AQ/CQ)

Because α, φ, γ are small, we get

OQ ≈ OP = u
IQ ≈ IP = v
CQ ≈ CP = r

Hence,

1/u + 1/v = 2/r

or,

1/u + 1/v = 1/f

Note: This formula is also applicable to a convex mirror.

26.3.2 Linear Magnification of Curved Mirrors

1. m in terms of u and v

The linear magnification m of the image produced by a curved mirror is defined as

m = v / u

where v is the image distance from the mirror
u is the object distance from the mirror

2. m in terms of h_i and h_o

The linear magnification m is also defined as

m = h_i / h_o

where h_i is the height of the image
h_o is the height of the object

26.3.3 Sign Convention for Mirror Formulae

(a) For u and v

Before we apply the two formulae given above, we need to adopt a sign convention. We are going to use the sign convention of real is positive.

It means that if the object and image are real, then the values of u and v are positive. Conversely, if the object and image are virtual, then the values of u and v are negative.

(b) For r, f

The values of r and f are positive for a concave mirror. Conversely, the values of r and f are negative for a convex mirror.

EXAMPLE 26.2

An object is placed at a distance of 30 cm from a spherical mirror of focal length 20 cm. Determine the position and also state the nature of the image produced if the mirror is (a) concave (b) convex. Sketch a ray diagram for the convex mirror.

Answer

(a)

Given

u = +30 cm (real object)
f = +20 cm (concave mirror)

1/f = 1/u + 1/v
1/20 = 1/30 + 1/v

v = +60 cm (real image)

m = v/u = 60/30 = 2 (magnified)

Nature of image:

• (i) real
• (ii) inverted
• (iii) magnified
• (iv) in front of the mirror

(b)

f = −20 cm (convex mirror)

1/(-20) = 1/30 + 1/v

v = −12 cm (virtual image)

m = v/u = -12/30 = -0.4 (diminished)

Nature of image:

• (i) virtual
• (ii) upright
• (iii) diminished
• (iv) behind the mirror

Ray diagram:

Figure 26.13 This figure shows the ray diagram for a convex mirror, illustrating how reflected rays diverge and appear to originate from a virtual image formed behind the mirror.

Figure 26.13 shows the ray diagram for the convex mirror.

EXAMPLE 26.3

An upright image of an object is formed by a spherical mirror at a distance of 30 cm from the real object. Its height is twice that of the object. Determine (a) the position of the mirror from the object (b) the radius of curvature of the mirror. State whether the mirror is concave or convex.

Answer

Figure 26.14 Ray diagram illustrating an upright virtual image formed behind the mirror, where reflected rays appear to diverge from the image position.

(a)

Because it is upright, the image is virtual. Hence, it has to be formed behind the mirror, as shown in Figure 26.14.

Given

m = 2 = v/u
v = 2u

Referring to the diagram, we get

v = 30 − u
2u = 30 − u
u = +10 cm

The mirror is 10 cm from the object.

v = −(30 − u)    (virtual image)
v = −30 + 10 = −20 cm

1/f = 1/u + 1/v
= 1/10 + 1/(-20) = 1/20

f = +20 cm

The mirror is concave.

r = 2f
= 2(+20) = +40 cm

EXAMPLE 26.4

A small object is placed upright in front of a concave mirror. A real image whose height is twice that of the object is formed. When the object is placed at another position, a virtual image whose height is also twice that of the object is formed. Determine the distance between the two images.

Answer

Real image:

Figure 26.15 (a) Ray diagram illustrating a real, inverted, and magnified image formed by a concave mirror.

Refer to Figure 26.15(a). Given

m₁ = 2 = v₁/u₁
u₁ = 1/2 v₁
f = 1/2 r

1/f = 1/u + 1/v
2/r = 2/v₁ + 1/v₁ = 3/v₁

v₁ = 3/2 r

Virtual image:

m₂ = 2 = v₂/u₂
u₂ = 1/2 v₂

2/r = 1/u₂ + 1/v₂
2/r = 2/u₂ + 1/(-u₂) = 1/u₂

v₂ = 1/2 r

Distance between the two objects = v₁ + v₂

= 3/2 r + 1/2 r = 2r

Figure 26.15(b) Ray diagram illustrating a virtual, upright, and magnified image formed by a concave mirror.

EXAMPLE 26.5

A straight rod is 10 cm long. It is placed with its length on the principal axis of a concave mirror of radius of curvature 30 cm. The end of the object which is nearer to the mirror is 18 cm from the mirror. Determine the magnification of the image of the rod.

Answer

Figure 26.16 Ray diagram showing image formation of a rod along the principal axis, where different points form images at different distances, leading to magnification.

Refer to Figure 26.16. Let points A and B represent the two ends of the rod.

Point A:

f = 1/2 r
= 1/2 (30) = 30/2 cm

u_A = 18 cm

1/f = 1/u_A + 1/v_A
1/v_A = 2/30 − 1/18 = 1/90

v_A = 90.0 cm

Point B:

Applying the same method to point B, we get

v_B = 32.3 cm

The length L_i of the image of the rod is

L_i = v_A − v_B
= 90.0 − 32.3 = 57.7 cm

The linear magnification of the image is

m = L_i / L_o
= 57.7 / 10 = 5.8

EXAMPLE 26.6

A plane mirror is placed vertically 10 cm in front of a convex mirror so that it blocks half of the curved mirror. A pin is placed vertically 25 cm in front of the plane mirror. The images formed by the plane mirror and the convex mirror are at the same position. Determine the focal length of the convex mirror.

Answer

Figure 26.17 Diagram showing a pin, a plane mirror, and a convex mirror where both mirrors produce images at the same position.

Refer to Figure 26.17

Plane mirror: Since the pin is 25 cm in front of the plane mirror, the image formed by it is 25 cm behind the plane mirror.

Convex mirror: The image formed by the convex mirror is formed at distance

u = −(25 − 10) = −15 cm   (virtual image)
u = 25 + 10 = 35 cm

1/f = 1/u + 1/v
1/f = 1/35 + 1/(−15) = −4/105

f = −26.3 cm
  

26.3.4 Virtual Object

Figure 26.18 Ray diagram illustrating how converging light reflected by a convex mirror forms a real image, with a virtual object located behind the mirror.

1. Consider a beam of light converging towards a convex mirror, as shown in Figure 26.18. The light beam would have converged to a point behind the mirror. However, it is reflected and converges instead at one point in front of the mirror, thus forming a real image I there.
2. The object for this real image I, the point found behind the mirror at R, then acts as a virtual object.
3. If we adopt the sign convention of ‘real is positive’ then the virtual object would have a negative value.

Example 26.7

Explain how a real image can be formed at a distance of 20 cm in front of a convex mirror, which has a radius of curvature of 30 cm. Determine the position of the virtual object for the real image.

Answer

Figure 26.19 Ray diagram showing how a convex mirror forms a real image 20 cm in front of the mirror when illuminated by converging light.

Shine a converging light beam onto the convex mirror. Adjust the position of the mirror until a real image is formed 20 cm in front of the mirror, as shown in Figure 26.19.

v = +20 cm          (real image)

f = 1/2 r
  = 1/2 (−30)
  = −15 cm          (convex mirror)

1/f = 1/u + 1/v

1/(−15) = 1/u + 1/(+20)

1/u = −1/15 − 1/20
    = −7/60

u = −8.6 cm         (virtual object)

Real-Life Applications

• Optical Instruments: Used in telescopes and microscopes to magnify images.
• Vehicle Mirrors: Convex mirrors provide a wider field of view for drivers.
• Light Reflectors: Concave mirrors focus light in headlights and flashlights.
• Makeup Mirrors: Concave mirrors produce enlarged images for detailed viewing.
• Medical Tools: Used by doctors to focus light during examinations.

Example: A car uses a convex mirror so the driver can see more area behind, even though the image appears smaller.

Conclusion

The relationship between object distance (u), image distance (v), and focal length (f) is fundamental in understanding how curved mirrors form images.

By applying the mirror formula and magnification concept, we can determine the position, size, and nature of images in various optical systems.

These principles are widely used in real-life applications such as optical instruments, vehicle mirrors, and lighting systems.

Mastering these concepts helps build a strong foundation in physics and optics.

Frequently Asked Questions

What is the mirror formula?

The mirror formula is 1/f = 1/u + 1/v. It relates focal length, object distance, and image distance.

What is magnification?

Magnification is the ratio of image size to object size and is given by m = v/u.

What is the difference between concave and convex mirrors?

A concave mirror can form real and virtual images, while a convex mirror only forms virtual and smaller images.

Why are convex mirrors used in vehicles?

Convex mirrors provide a wider field of view, helping drivers see more area behind them.

When is an image real or virtual?

A real image is formed when light rays actually meet, while a virtual image is formed when rays appear to meet.

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