Thin Lenses: Formula, Ray Diagrams, Applications & Solved Examples

26.5 THIN LENSES

26.5.1 Types of Thin Lenses

Figure 26.24(a) shows three possible shapes of a thin convex lens while Figure 26.24(b) shows three possible shapes of a thin concave lens.

Figure 26.24 (a) shows three possible shapes of a thin convex lens, which are thicker at the center than at the edges and converge light rays. (b) shows three possible shapes of a thin concave lens, which are thinner at the center than at the edges and diverge light rays.

26.5.2 Some Term Concerning Thin Lenses

(a) Centre and radius of curvature

Figure 26.25 Each surface of a lens is spherical in shape unless it is flat. The first surface has a centre of curvature C₁, and the second surface has a centre of curvature C₂. These centres define the curvature of the lens surfaces and are essential in analyzing image formation through the lens.

Each surface of a lens is spherical in shape, unless it is flat. Hence, one surface has a centre of curvature C₁ and the other surface has a centre of curvature C₂, as shown in Figure 26.25.

(b) Radius of curvature

The radius of each spherical surface is known as the radius of curvature.

(c) Principal axis

The line which passes through C₁ and C₂ is known as the principal axis.

26.5.3 Focus and Focal Length

Figure 26.26 (a) A convex lens refracts parallel light rays so that they converge at a point F on the principal axis. This point is called the focus, and the distance OF is the focal length. (b) A concave lens refracts parallel light rays so that they diverge. The point F is the apparent origin of the rays and is called the focus, while OF represents the focal length.

(a) Convex lens

Consider a narrow light beam travelling along the principal axis and incident upon a thin convex lens, as shown in Figure 26.26(a). The light beam will refract when it enters the lens. After emerging from the lens, the beam converges to a point F on the principal axis. This point is known as the focus. The vertical plane AB on which it lies is the focal plane. The length OF between the centre of the lens and the focus is the focal length of the lens.

(b) Concave lens

Replace the convex lens with a thin concave lens, as shown in Figure 26.26(b). After emerging from the concave lens, the light beam will diverge. The point F which appears to be the origin of the diverging beam is the focus of the concave lens. The length OF between the centre of the lens and the focus is the focal length of the lens.

26.5.4 Ray Diagram

Figure 26.27 shows the three principal rays used to construct ray diagrams for a thin lens.
(a) A ray passing through the optical centre O travels in a straight line without deviation.
(b) A ray parallel to the principal axis is refracted through the focus F on the opposite side of the lens.
(c) A ray directed towards the focus F emerges from the lens parallel to the principal axis.

To construct a ray diagram of a thin lens, we normally make use of three ‘special rays’. We draw these rays as follows:

Ray 1: Because the lens is thin, we normally draw a straight line through the centre O of the lens, as shown in Figure 26.27(a), to represent a light ray through O.

Ray 2: Draw a line parallel to the principal axis until it touches the lens. Let it pass through the focus F on the opposite side of the lens, as shown in Figure 26.27(b).

Ray 3: Draw a line passing through the focus F and allow it to come to the lens. After emerging from the lens, draw a line parallel to the principal axis, as shown in Figure 26.27(c).

26.5.5 Using Ray Diagram to Deduce the Nature of the Image Formed by Thin Convex Lens

(a) Object distance longer than 2f

Figure 26.28 When an object is placed at a distance greater than 2f from a convex lens, the image formed is inverted, real, and diminished in size. The image is located between f and 2f on the opposite side of the lens.

Refer to the ray diagram shown in Figure 26.28. The image of the object is:

  • (i) inverted
  • (ii) real
  • (iii) diminished
  • (iv) on the opposite side of the lens

(b) Object distance = 2f

Figure 26.29 When the object is placed at 2f from a convex lens, the image formed is inverted, real, and the same size as the object. The image is located at 2f on the opposite side of the lens.

Refer to the ray diagram shown in Figure 26.29. The image of the object is:

  • (i) inverted
  • (ii) real
  • (iii) same size as the object
  • (iv) on the opposite side of the lens

(c) Object distance between f and 2f

Figure 26.30 When the object is placed between f and 2f, the image formed is inverted, real, and magnified. The image is located beyond 2f on the opposite side of the lens.

Refer to the ray diagram shown in Figure 26.30. The image of the object is:

  • (i) inverted
  • (ii) real
  • (iii) magnified
  • (iv) on the opposite side of the lens

(d) Object distance = f

Figure 26.31 When the object is placed at the focal point f, the refracted rays emerge parallel and the image is formed at infinity.

Refer to the ray diagram shown in Figure 26.31. The image of the object is at infinity.

(e) Object distance less than f

Figure 26.32 When the object is placed at a distance less than f, the image formed is upright, virtual, and magnified. The image is located on the same side of the lens as the object.

Refer to the ray diagram shown in Figure 26.32. The image of the object is:

  • (i) upright
  • (ii) virtual
  • (iii) magnified
  • (iv) on the same side of the lens as the object

26.5.6 Using Ray Diagram to Deduce the Nature of the Image Formed by Thin Concave Lens

Figure 26.33 For a concave lens, regardless of the object position, the image formed is upright, virtual, and diminished in size. The image is located on the same side of the lens as the object.

Refer to the ray diagram shown in Figure 26.33. The image of the object is:

  • (i) upright
  • (ii) virtual
  • (iii) diminished
  • (iv) on the same side of the lens as the object

26.5.7 Virtual Object

Figure 26.34 (a) shows a beam of light that is about to converge at point O. When a convex lens is placed in the path of the beam, the rays are refracted and converge at a new point I. (b) shows that a similar effect occurs with a concave lens, where the incoming converging rays are refracted and form an image at point I. In both cases, point O acts as a virtual object for the image formed at I.

Consider a beam of light which is about to converge at point O, as shown in Figure 26.34(a). If we intercept the light beam with a convex lens, then the beam will now converge at point I instead. The same thing will happen if we use a concave lens, as shown in Figure 26.34(b).

For each lens, a real image I is formed. The object for this image is the point O, which does not have light passing through it. Hence, O is a ‘virtual’ point. We call O the virtual object for image I.

26.5.8 Refraction at Two Spherical Surfaces

Figure 26.35 shows a thin convex lens of refractive index n₂ immersed in a medium of refractive index n₁ (n₂ > n₁). A point object O is placed at a distance u from the lens. Light from the object is first refracted at the surface PP′ (radius r₁), forming an intermediate image I′ if the second surface is ignored. The ray then passes through the second surface QQ′ (radius r₂), where it is refracted again and finally forms an image I at distance v. The intermediate image I′ acts as a virtual object for the second refraction.

  1. A thin convex lens of refractive index n₂ is totally immersed in a transparent medium of refractive index n₁ where n₂ > n₁.
  2. A point object O is inside the transparent medium and is placed at distance u from a convex lens, as shown in Figure 26.35(a). The surface PP′ facing the object has radius of curvature r₁.
  3. A ray of light from the object is refracted as it passes through PP′. Not long after that, the ray passes through the second surface QQ′ of the lens and is refracted further. QQ′ has radius of curvature r₂.
  4. The ray emerged from the lens and forms a point image at distance v from the lens.
  5. The lengths u, v, r₁ and r₂ are related to one another by the formula
1/u + 1/v = (n₂/n₁ − 1)(1/r₁ + 1/r₂)

Derivation:

Refraction at first surface PP′:

Figure 26.35(b) shows the ray of light from the object O being refracted at the first surface PP′ of the lens. Suppose that the second surface QQ′ does not exist and the space to the right of PP′ is completely filled with the medium of refractive index n₂. Then the ray would form a real point image I′ at distance v′ from the surface PP′. For refraction on PP′, we have

n₁/u + n₂/v′ = (n₂ − n₁)/r₁

Refraction at second surface QQ′:

Let us bring back the second surface QQ′ of the lens. This surface will intercept the ray so that the ray cannot form the image I′. Instead, the ray undergoes refraction at QQ′ and finally emerges into the medium of refractive index n₁. Now the ray forms a real point image I at distance v from QQ′, as shown in Figure 26.35(c).

Image I is real but the point object for this image is virtual. The virtual object is I′, which is supposed to be formed at distance v′ from PP′.

If the lens is thin, then its thickness may be neglected. We can assume that distance of virtual object I′ from PP′ = distance of I′ from QQ′ = −v′ (virtual object)

distance of real image I from QQ′ = v

The virtual object is in medium with n₂ while the real image is in medium with n₁. Hence, for the refraction on QQ′ we have

n₁/u + n₁/v = (n₂ − n₁)/r₂     (2)
n₁/u + n₁/v = (n₂ − n₁) (1/r₁ + 1/r₂)

Dividing by n₁,

1/u + 1/v = (n₂/n₁ − 1) (1/r₁ + 1/r₂)

26.5.9 The Lens Maker’s Formula

Refer to Figure 26.35(a). If the object is very far away from the lens, the image of the object will be formed at the focus. Hence, we have

u = ∞ ,   v = f
1/∞ + 1/f = (n₂/n₁ − 1) (1/r₁ + 1/r₂)
1/f = (n₂/n₁ − 1) (1/r₁ + 1/r₂)

This equation is known as the lens maker’s formula. Suppose that the lens is in air with n₁ = 1 and the refractive index of the lens is n. Then the equation above becomes

1/f = (n − 1) (1/r₁ + 1/r₂)     (lens in air)

EXAMPLE 26.10

Figure 26.36 The diagram shows a thin lens with two curved surfaces P and Q, used to determine its focal length and lens type in different media.


The shape of a thin lens is shown in Figure 26.36. The radius of curvature of surfaces P and Q are 20 cm and 50 cm respectively. The refractive index of the lens is 1.52. Determine the focal length of the lens if the lens is (a) in air (b) in water which has a refractive index of 1.33. Does the lens act as a convex or concave lens?

Answer

(a) Given

rₚ = +20 cm,   r_Q = −50 cm,   n₁ = 1.0,   n₂ = 1.52
1/f = (n − 1) (1/r₁ + 1/r₂)
= (1.52 − 1) (1/20 + 1/(−50)) = (0.52)(3)/100
f = +64.1 cm

The lens is a convex lens.

(b) Given

rₚ = +20 cm,   r_Q = −50 cm,   n₁ = 1.33,   n₂ = 1.52
1/f = (n₂/n₁ − 1) (1/r₁ + 1/r₂)
= (1.52/1.33 − 1) (1/20 + 1/(−50))
f = +233 cm

The lens is a convex lens.

26.5.10 Formula for Thin Lens

For a thin lens, we have

1/u + 1/v = (n₂/n₁ − 1) (1/r₁ + 1/r₂)

and

1/f = (n₂/n₁ − 1) (1/r₁ + 1/r₂)

Therefore,

1/f = 1/u + 1/v

26.5.11 Sign Convention for Thin Lens

Before we use the thin lens formula, we need to adopt a sign convention. We are going to use the sign convention of ‘real is positive’.

(a) For real objects and images, the object distance and image distances have positive values.

(b) For a convex lens, the focal length has a positive value. For a concave lens, the focal length is negative.

EXAMPLE 26.11

A thin plano-concave lens has one flat surface and one spherical surface with radius of curvature of 15 cm. It has refractive index 1.50. An object is placed 20 cm from the lens. Determine

(a) the focal length of the lens
(b) the image distance from the lens.

Answer

(a) Given

n = 1.50,   r₁ = ∞,   r₂ = −15 cm,   u = +20 cm
1/f = (n − 1) (1/r₁ + 1/r₂)
= (1.50 − 1) (1/∞ + 1/(−15))
= −1/30
f = −30 cm

(b)

1/f = 1/u + 1/v
1/(−30) = 1/20 + 1/v
1/v = 1/(−30) − 1/20
v = −12 cm

EXAMPLE 26.12

Two thin convex lenses, each of focal length 10 cm, are placed on one common axis. A point source, also on the axis, is placed 4.0 cm from one of the lenses. Determine the separation of the lenses if light emerges from the second lens as a parallel beam.

Answer

Refer to the ray diagram shown in Figure 26.37.

Figure 26.37 The diagram shows two thin convex lenses placed on a common axis with a point source near the first lens. It is used to determine the separation between the lenses such that the light emerging from the second lens forms a parallel beam.

Consider lens L₂ only:

The ray emerges from L₂ as a parallel ray. Hence, the ray appears to originate from point I′ at distance v′ from L₂. In other words, I′ has to be the focus of L₂.

f = v′ + d

Consider lens L₁ only:

I′ is the virtual image of the point object formed by light passing through L₁. Hence, for lens L₁, we have

f = +10 cm
u = +4.0 cm
v′ = distance of virtual image I′
1/f = 1/u + 1/v′
1/10 = 1/4.0 + 1/v′
v′ = −6.67 cm

Referring back to lens L₂, we have

10 = |−6.67| + d
d = 3.3 cm

EXAMPLE 26.13

Figure 26.38 The diagram shows a thin convex mirror A, a point source B, a convex lens C, and a screen arranged along a common axis. Light from B is reflected by the mirror, refracted by the lens, and forms a sharp image on the screen.

A thin convex mirror A, a point source B, a thin convex lens C of focal length 15 cm, and a screen are arranged in such a way that an axis passes through them, as shown in Figure 26.38. A sharp image is formed when light from B is reflected by A, refracted by C and finally incident on the screen. Determine the radius of curvature of the mirror.

Answer

Figure 26.39 The diagram shows that the image formed at point I₁ behaves as an object for the lens. When considering the lens alone, the light appears to originate from I₁, which acts as the object for further image formation.

Refer to the ray diagram shown in Figure 26.39. If we only look at the lens, we will notice that a light source appears to be at I₁. Hence, I₁ acts as an object.

u₁ = distance of I₁ from the lens
v₁ = +25 cm
f₁ = +15 cm
1/f₁ = 1/u₁ + 1/v₁
1/+15 = 1/u₁ + 1/+25
u₁ = +37.5 cm

Next, consider the mirror only. Now I₁ is the virtual image of the point source B. We have

u₂ = +20 cm
v₂ = −(u₁ − 20 − 10)
= −(37.5 − 30) = −7.5 cm (virtual image)
1/f₂ = 1/u₂ + 1/v₂
= 1/+20 + 1/(−7.5)
f₂ = −12 cm
radius of curvature = 2f₂
= 2(−12)
= −24 cm

26.5.12 Graph of u against v for Thin Convex Lens

1 If we plot u against v for a thin lens, we will get two hyperbolae, as shown in Figure 26.41(a). The curves are for the following:

Curve A: for both real objects and real images.
Curve B: for virtual objects but real images.
Curve C: for real objects but virtual images.

Figure 26.41(a) shows the graph of object distance u against image distance v for a thin lens, forming hyperbolic curves. Curve A represents real objects and real images, Curve B represents virtual objects and real images, and Curve C represents real objects and virtual images. (b) shows the line u = v plotted on the same graph. The intersection points (u₀, v₀) indicate the condition where object distance equals image distance.

2 Figure 26.41(b) shows the line for the expression u = v being plotted on the u against v graph. The line cuts the curve at points u₀ and v₀. Since we have u = v, then

(a) u₀ = v₀

(b) 1/f = 1/u₀ + 1/v₀
= 2/u₀
u₀ = 2f

26.5.13 Graph of u + v Against u for Thin Convex Lens

Figure 26.42 The graph shows the relationship between (u + v) and u (or v) for a thin convex lens based on the lens formula. The curve reaches a minimum value when u = v, at which point (u + v) = 4f.

For a thin convex lens, we have

1/f = 1/u + 1/v
= (u + v) / uv

Figure 26.42 shows how a graph of u + v plotted against u (or v) looks like. It can be shown that the minimum point of the curve occurs at

u + v = 4f

This occurs when u = v.

26.5.14 Linear Magnification by Thin Lenses

1 Definition

The linear magnification m of an image produced by a thin lens is defined as the ratio

m = v/u = hᵢ / h₀

where u is the object distance
v is the image distance
hᵢ is the height of the image
h₀ is the height of the object

2 m in terms of f and v

For a thin lens, we have

1/f = 1/u + 1/v
v/f = v/u + 1
v/f = m + 1

3 Graph of v against m

Figure 26.43 The graph shows a linear relationship between image distance v and magnification m for a thin lens. The intercept on the v-axis represents the focal length f of the lens.


If we plot v against m for a thin lens, we will get a line, as shown in Figure 26.43. The intercept on the v-axis is equal to the focal length f.

EXAMPLE 26.14

Figure 26.44 The diagram shows ray constructions for a thin convex lens where an object forms a real, inverted image with twice its size and a virtual, upright image also magnified by a factor of two. It is used to determine the object positions required to produce both types of images.

An object is placed 15 cm from a thin convex lens. The lens produces a real image whose size is twice that of the object. At what distance from the lens must the object be placed in order to obtain a virtual image whose size is also twice that of the object? Sketch a ray diagram for each image.

Answer

(a) Real image

u₁ = +15 cm
m₁ = 2
m₁ = v₁ / u₁
v₁ = m₁ u₁
= (2 × 15)
= +30 cm

(real image)

1/f = 1/u₁ + 1/v₁
= 1/+15 + 1/+30
= 1/10
f = +10 cm

(b) Virtual image

m₂ = 2 = v₂ / u₂
v₂ = −2u₂
1/f = 1/u₂ + 1/v₂
= 1/u₂ + 1/(−2u₂)
= 1/(2u₂)
u₂ = 1/2 f
= 1/2 (+10)
= +5 cm

Figure 26.44 shows the rays diagrams for the two images.

EXAMPLE 26.15

Determine the focal length of a thin convex lens which produces a virtual image whose size is five times that of the object when the object is placed 20 cm from the lens.

Answer

u = +20 cm
m = v/u = 5
v = −5u
= −5(20)
= −100 cm

(virtual object)

1/f = 1/u + 1/v
1/f = 1/20 + 1/(−100)
= 1/25
f = +25 cm

🔬 Applications of Thin Lenses

👓 Eyeglasses

Thin lenses are used to correct vision problems. Convex lenses help with long-sightedness, while concave lenses correct short-sightedness.

📷 Cameras

Convex lenses focus light onto a sensor to produce clear images. Adjusting lens position allows focusing at different distances.

🔬 Microscopes

Multiple convex lenses are used to magnify very small objects, producing enlarged images for scientific study.

🔭 Telescopes

Lenses collect and focus light from distant objects like stars and planets, making them visible and clearer.

💡 Projectors

Convex lenses are used to project enlarged real images onto screens in classrooms, cinemas, and presentations.

🔍 Magnifying Glass

A simple convex lens that produces a magnified virtual image when the object is placed within its focal length.

✨ Conclusion

Thin lenses are essential in understanding how light forms images through refraction. By applying the thin lens formula, we can determine the position, size, and nature of images formed by convex and concave lenses.

This concept is widely used in everyday life, from eyeglasses and cameras to microscopes and telescopes. It plays a key role in modern technology, optical instruments, and scientific advancements.

Mastering thin lens concepts not only strengthens problem-solving skills in physics but also provides a deeper understanding of how visual systems and optical devices work.

❓ Frequently Asked Questions (FAQ)

A thin lens is a lens whose thickness is very small compared to its radius of curvature, allowing us to simplify calculations using the thin lens formula.

The thin lens formula is 1/f = 1/u + 1/v, where f is focal length, u is object distance, and v is image distance.

Convex lenses converge light rays and can form real images, while concave lenses diverge light rays and typically form virtual images.

A real image is formed when light rays actually meet, while a virtual image is formed when rays appear to come from a point but do not actually meet.

Thin lenses are used in eyeglasses, cameras, microscopes, telescopes, projectors, and many optical devices.

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