The behavior and characteristics of light are studied in optics, a branch of physics. Ray Optics, also known as Geometrical Optics, discusses that light travels in a straight path and that every object has an image. Optical instruments are optical devices that have been created using the reflecting and refracting capabilities of mirrors, lenses, and prisms.
What is a Lens?
Lenses are basically magnifying lenses with curved sides. A lens is a piece of transparent glass that focuses or disperses light beams through refraction. Because of their magnifying properties, lenses are utilized in telescopes and other magnifying tools. They capture light beams and are utilized in cameras.
Light is gathered by a group of lenses rather than a single lens in cameras. The magnification of a lens is the proportion of the picture generated to the size of the target. Lenses can also be used in groups to lessen blurriness and distortion in the images they produce.
As a result, a lens is a tiny piece of material that has been changed to capture a picture using light rays. In this post, we’ll go through the lens formula, the terminologies associated with it, and how to use the lens formula to calculate magnification.
Terminology used for Lenses
- Pole –The pole is the center of the lens’s spherical refracting surface. The point at which the primary axis intersects the lens’s surface.
- Optical Centre – The optical center is the point on the principal axis at the center of the lens.
- Centre of Curvature – A lens has two spherical surfaces, which together form a sphere. The center of curvature is the point at which these spheres intersect.
- Principal axis – The principal axis is an imaginary line that connects the curvature center and the pole.
- Aperture – Aperture refers to the portion of the lens that is suitable for refraction. The lens’s aperture is the effective diameter of its light-transmitting region.
- Focus – The point at which collimated light parallel to the axis is concentrated is referred to as the focus.
- The focal length – The focal length of a lens is the distance between its optical center and its focal point or focus.
- Power – The lens’s power is the reciprocal of its focal length. Dioptre is the SI unit of power.
Types of Lenses
A simple lens is a single magnifying component, but a compound lens is made up of several simple lenses aligned along a common axis. Optical aberrations can be seen in simple lenses, however, they are not present in compound lenses. Another benefit of the compound lens is that the magnification may be altered to meet the needs of the user.
They are divided into two groups based on the form and purpose of the lens:
- Concave lens
A concave lens is one that has at least one side that curves inwards. A biconcave lens is a concave lens with both sides curled inward. Concave lenses are divergent lenses, which means that light rays refracted through them are stretched apart. They possess the capacity to diverge a parallel light beam. The margins of a concave lens are either broader than the centre or narrower than the centre.
A concave lens gives the viewer a smaller picture. After passing through the lens, rays of light parallel to the axis seem to disperse, which is the focal point of a concave lens. The distance between the lens’s optical center and the focal point is known as focal length.
- Convex lens
A lens having an outward curvature is known as a convex lens. In contrast to a concave lens, the thickness of a convex lens is greater in the centre than it is at the corners. Converging lenses are known as convex lenses. They can concentrate a parallel stream of light into a single point. The focus point of a convex lens is this location, and the focal length is the distance between the optical centre and the focal point. The focal point is located on the opposite side of the lens than the light beams.
A Plano-convex lens is a convex lens with one side flat. The human eye’s lens is an excellent example of a convex lens. A magnifying glass, which is used to treat Hypermetropia or long-sightedness, is another example of a convex lens. In cameras, convex lenses are used to concentrate light and provide a clean image. Convex lenses are also seen in compound lenses used in magnification equipment like microscopes and telescopes.
Combination of Thin Lenses in Contact
Consider two lenses, M and N with focal lengths of f1 and f2, respectively. Object S is placed on the common principal axis. The first lens M creates an image at I, which serves as the subject for the second lens N. As seen in the figure, the final picture is created at I1.
Let object distance for the first lens (M), PS be u, the final image distance, PI be v and the Image distance for the first lens (M), PI1 be v.
Now, for the image I1 produced by the first lens M, the lens formula can be written as:
1/v1− 1/u = 1/f1 …. (1)
For the final image I, produced by the second lens N,
1/v − 1/v1 = 1/f2 …. (2)
Adding equations (1) and (2),
1/v − 1/u = 1/f1 + 1/f2 …. (3)
If the combination is replaced by a single lens of focal length F such that it forms the image of S at the same position I, then
1/v − 1/u = 1/F …. (4)
From equations (3) and (4),
1/F = 1/f1 + 1/f2
- where F is the focal length for the equivalent lens for the combination.
If n numbers of lenses are present, the equation can be written as:
1/F = 1/f1 + 1/f2 + ……. + 1/fn
But what happens if the lenses aren’t in contact?
The combined focal length may be determined using the following formula if the lenses are separated by a distance ‘d’.
1/F = 1/f1 + 1/f2 + ……. + d/fn
Applications of Combination of Lenses
A combination of Lenses is used in telescopes and microscopes to combine two or more lenses to:
- Obtain an object’s erect image.
- By using a single lens, you can reduce the number of defects that occur.
- Increase the image’s magnification.
Sample Problems
Problem 1: A convex lens with a focal length of 9 cm interferes with a concave lens with a focal length of 19 cm. What is this lens combination’s focal length?
Solution:
Given that,
f1 = +9 cm for convex lens,
f2 = −19 cm for concave lens.
The focal length of this combination is given by,
1/F = 1/f1 + 1/f2
1/F = 1/9 + 1/-19
1/F = 10/171
F = +17.1 cm
Hence, the total focal length of this combination is +17.1 cm.
Problem 2: A distance of 8 cm separates two narrow convex lenses with focal lengths of 12 cm and 17 cm. The combination’s focal length is?
Solution:
f1 = 12 cm,
f2 =17 cm
d = 8 cm.
The focal length of this combination is F, is given by
1/F = 1/f1 + 1/f2 – d/f1.f2
1/F = 1/12 + 1/17 – 8/12 × 17
= 1/12 + 1/17 – 2/51
= 7/68
F = 68/7
= 9.71 cm
Hence, the total focal length of this combination is 9.71 cm.
Problem 3: What is the focal length of a concave lens with a focal length of 16cm in contact with a convex lens with a focal length of 28 cm? Is the system’s lens converging or diverging? Ignore the lens thickness.
Solution:
Given,
The Focal length of convex lenses,
f1 = 28 cm
f2 = 16 cm
The Focal length of concave lenses,
f2 = -16 cm
From the formula,
1/f = 1/f1 + 1/f2
1/f = 1/28 – 1/16
f = -112/3
Substitute to get the answer as,
f = -37.3 cm
Hence, it is a diverging lens.
Problem 4: Two power lenses, -19 D and +6 D are in contact with each other. What is the focal length of the combination of these lenses?
Solution:
We know that,
Power, P = 1/f
P = P1 + P2
1/f = 1/f1 + 1/f2
1/f = -19 + 6
f = 1 / -13 m
= – 0.076 m
= – 7.6 cm
Problem 5: Five converging lenses of identical focal length ‘f’ are brought together. Determine the combination’s focal length.
Solution:
Let,
f1 = f2 = f3 = f4 = f5 = f
So the combinational focal length be f
from formula,
1/F = 1/f1 + 1/f2 + 1/f3 + 1/f4 + 1/f5
= 1/f + 1/f + 1/f + 1/f + 1/f
= (1+1+1+1+1) / f
1/F = 5/f
Hence,
F = f/5
Problem 6: Two converging and two diverging lenses with similar focal lengths are positioned coaxially in contact. Determine the combination’s focal length and power.
Solution:
as we know the formula,
1/F = 1/f1 + 1/f2 + 1/f3 + 1/f4 + 1/f5
f1 = +f (for converging lens)
f2 = +f (for converging lens)
f3 = -f (for diverging lens)
f4 = -f (for diverging lens)
Now substitute,
1/F = 1/f +1/f – 1/f – 1/f
1/f = 2/f – 2f
F = 1/0
= infinity (∞)
Since,
P = 1/F
= 1/ infinity (∞)
= 0
Hence, the power of this combination lens is 0.
Problem 7: Derive the relation 1/f = 1/f1 + 1/f2, focal lengths of two thin lenses are f1 and f2 and f is the focal length of the combination in contact, and if the focal length of two lenses is 15.5 cm and 17.5 cm, calculate the total focal length of this combination.
Solution:
Consider two thin lenses in contact of focal length f1 and f2.
For lens 1,
1/v1 – 1/u = 1/f1 ….(a)
for lens 2,
1/v -1/v1 = 1/f2 …..(b)
Now add equation (a) and (b),
1/f1 + 1/f2 = 1/v – 1/u + 1/v1 -1/v1
1/f1 + 1/f2 = 1/v – 1/u
Now using lens formula: (1/v – 1/u = 1/f)
1/f1 + 1/f2 = 1/f
For thin lenses, n number of contact,
1/f = 1/f1 + 1/f2 + 1/f3 +……….+ 1/fn
Hence derived.
It is given that:
f1 = 15.5 cm
f2 = 17.5 cm
From the formula,
1/f = 1/f1 +1/f2
Now substitute the given values,
1/f = 1/15.5 + 1/17.5
1/f = 132/1085
f = 1085/132
= 8.21 cm
Therefore, the total focal length of this combination is 8.21 cm.