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Types of lenses

A lens is a transparent medium bounded by two surfaces of regular shape. There are two major categories of lenses which include:

Convex lens -they are thicker at the middle than at the edges.
Concave lens- they are thinner at the middle than at the edges.

Ray diagrams and terms used

Optical center is a geometric center of a lens.

Center of curvatureis the center of the sphere in which a lens is a part.

Principal axisis an imaginary line which passes through the optical center of the lens at right angle to the lens.

Principle focusis a point through which all rays traveling close and parallel to the principal axis pass through.

Images formed: Ray construction

Rays diagrams are normally used toillustratesthe formation of images by lenses.

A ray parallel to the principal axis passes through or appears to diverge from the principal focus after refraction.
A ray of light passing through the principal focus of a lens is refracted parallel to the principal axis of the lens.
A ray of light through the optical center of the lens continues throughundeviated(Not change direction)

Images formed: Characteristics

The nature, position and size of the image formed by a lens depends on the position of the object in relation to the type of lens. For example in converging lens when the object is between the lens and principal focus the image will be formed at the same side as the object but further from the lens. It is virtual, erect, and magnified. The image by concave lens is erect, virtual and reduced

The nature, position and relative size of the image formed by convex lens for various positions of the object is summarized in the table below:

Position of the object	Position of the image	Relative size of the image	Nature of the image
At infinity	At focus F₂	Highly diminished, point-sized	Real and inverted
Beyond 2F₁	Between F₂and 2F₂	Diminished	Real and inverted
At 2F₁	At 2F₂	Same size	Real and inverted
Between F₁and 2F₁	Beyond 2F₂	Enlarged	Real and inverted
At focus F₁	At infinity	Infinitely large or highly enlarged	Real and inverted
Between focus F₁and optical centre O	On the same side of the lens as the object	Enlarged	Virtual and erect

Activity 1

Take a convex lens. Find its approximate focal length in a way described in Activity 11.
Draw five parallel straight lines, using chalk, on a long Table such that the distance between the successive lines is equal to the focal length of the lens.
Place the lens on a lens stand. Place it on the central line such that the optical centre of the lens lies just over the line.
The two lines on either side of the lens correspond to F and 2F of the lens respectively. Mark them with appropriate letters such as 2F₁, F₁, F₂and 2F₂, respectively.
Place a burning candle, far beyond 2F₁to the left. Obtain a clear sharp image on a screen on the opposite side of the lens.
Note down the nature, position and relative size of the image.
Repeat this Activity by placing object just behind 2F₁, between F₁and 2F₁at F₁, between F₁and O. Note down and tabulate your observations.

Nature, position and relative size of the image formed by a concave lens for various positions of the object

Position of the object	Position of the image	Relative size of the image	Nature of the image
At infinity	At focus F₁	Highly diminished, point-sized	Virtual and erect
Between infinity and optical centre O of the lens	Between focus F₁and optical centre O	Diminished	Virtual and erect

Activity 2

Take a concave lens. Place it on a lens stand.
Place a burning candle on one side of the lens.
Look through the lens from the other side and observe the image. Try to get the image on a screen, if possible. If not, observe the image directly through the lens.
Note down the nature, relative size and approximate position of the image.
Move the candle away from the lens. Note the change in the size of the image. What happens to the size of the image when the candle is placed too far away from the lens.

Images formed: Magnification

As we have a formula for spherical mirrors, we also have formula for spherical lenses. This formula gives the relationship between object distance (u), image-distance (ν) and the focal length (f ). The lens formula is expressed as1/ν - 1/u = 1/f(8)

The lens formula given above is general and is valid in all situations for any spherical lens. Take proper care of the signs of different quantities, while putting numerical values for solving problems relating to lenses.

The magnification produced by a lens, similar to that for spherical mirrors, is defined as the ratio of the height of the image and the height of the object. It is represented by the letter m. If h is the height of the object and h′ is the height of the image given by a lens, then the magnification produced by the lens is given by,m = Height of the Image / Height of the object = h' / h(9

Magnification produced by a lens is also related to the object-distance u, and the image-distance ν. This relationship is given byMagnification (m ) = h' / h = ν / u(10)

Example 1

A concave lens has focal length of 15 cm. At what distance should the object from the lens be placed so that it forms an image at 10 cm from the lens? Also, find the magnification produced by the lens

Solution

A concave lens always forms a virtual, erect image on the same side of the object.

Image-distance v = –10 cm;

Focal length f = –15 cm;

Object-distance u = ?

Since, 1 /v - 1 / u = 1 / f

or, 1 / u = 1 / v - 1 / f

1 / u = 1 / -10 - 1 / (-15) = - 1 / 10 + 1 / 15

1 / u = (-3+2) / 30 = 1 / (-30)

or, u = - 30 cm.

Thus, the object-distance is 30 cm.

Magnification m = v/ u

m = -10 cm / -30 cm = 1 / 3 = +0.33

The positive sign shows that the image is erect and virtual. The image is one-third of the size of the object.

Determination of Focal length: Estimation method

Determination of Focal length: Lens formula

Human eye, defects( short sightedness and long sightedness only)

Optical devices: Simple microscope

A magnifying glass, an ordinary double convex lens with a short focal length, is a simple microscope. For example, reading lens and hand lens. When an object is placed nearer such a lens than its principal focus, i.e., between the principal focus and the lens, an image is produced that is erect and larger than the original object. The image is also virtual; i.e., it cannot be projected on a screen as can a real image.

The image formed by magnifying glass or simple microscope is virtual and erect object place between principal focus (F) and convex lens.

The normal district vision
The position of the lens is usually adjusted so that V is about 25cm, which is the shortest distance of distinct vision.

Using the equation of lens (Lens formula).

Magnication (M) of simple microscope

Magnification is the ratio of the image distance 'v' to the object distance 'u'.

Example 2

A simple microscope with lens of focal length 5cm is used to read division of a scale 0.5mm in size. How large will the division be seen through the simple microscope?

Data given

Focal length, f = 5cm
Required to find magnification, M

Soln:

From

M = (25/f + 1)

= (25/5+1)

=(5+1)

= 6

The magnification of lens = 6

Let the size of the object be ho and that of the image be hi. Then:

M = h1/H ……………(i)

H1 = 6h

The Height , h = (0.5mm)

H1 = 6 (0.5mm)

HI = 3mm

Hence, each division will appear to have a size of 3.0mm viewed through the simple microscope.

Parts of simple microscope

Optical devices: Compound microscope

A compound microscope is an optical instrument used to produce much greater magnification than that produced by simple microscope. The main features of a compound microscope includes two short-focus convex lenses, the objective lens, and the eyepiece.

Demonstration

The most commonly used microscope for general purposes is the standard compound microscope. It magnifies the size of the object by a complex system of lens arrangement.

It has a series of two lenses; (i) the objective lens close to the object to be observed and (ii) the ocular lens or eyepiece, through which the image is viewed by eye. Light from a light source (mirror or electric lamp) passes through a thin transparent object.

The objective lens produces a magnified ‘real image’ (first image of the object). This image is again magnified by the ocular lens (eyepiece) to obtain a magnified ‘virtual image’ (final image), which can be seen by eye through the eyepiece. As light passes directly from the source to the eye through the two lenses, the field of vision is brightly illuminated. That is why it is a bright-field microscope.

The object lens forms a real and inverted image I_Iof the object O ( the image is slightly magnified). The eyepiece lens acts as a magnifying glass for the first image II and produces a magnifical virtual image.

The object is placed just beyond the principal (fo) of the objective lens so that that the real image I, is formed inside the principal focus (F) of the eye piece. The eyepiece treats the real image I, as an object and then forms its magnified virtual image I2.

Magnification of a compound microscope: This isthe ratio of the image distance produced by a compound microscope to the object distance. The magnification produced by objective lens is v/u.

Where

V is the image distance

U is the object distance

The magnification given by the eyepiece is given by;

Me = 25/fe + 1

If the final image is formed at the least distance of distinct vision (V = 25cm).

Mc = M_om_e

Combine eqn (i) and (ii)

Then

Mc = (v/u) (25/fe+1)

The above formula shows that the final virtual image is formed at the least distance of distinct vision.

The uses of a compound microscope includes the following:

Used to magnify microorganism such as bacteria which cannot be seen by naked eyes.
Used in hospitals widely to detect microorganisms in specimens provided by patients. A specimen is a small amount that is taken for testing. Blood is an example of specimens. In hospitals microscopes can detect parasites such as plasmodium ssp (a causative agent for malaria) in blood specimen.

Example 3

A certain microscope consists of two converging lenses of focal length 10cm and 4cm for the objective and eyepiece, respectively. The two lenses are separated by a distance of 30cm. The instrument is focused so that the final image is at infinity. Calculate the position of the object and the magnification of the objective lens.

For the objective lens

I/U + I/V = I/Fo

Where

Fo = 10cm

The objective lens forms a real image of the object at the principal focus of the eyepiece.

Thus

V = (30 – 4)

= 26cm

Thus I/U + I/V = I/10

I/U + 1/26 = 1/10

1/U = (1/10 – I/26)

(I/U) -1 = (4/65)

(1/U) -1 = (4/65)-1

U = (65/4)

The magnification given by the objective lense is given by:

Whereas:

V = 26cm

U= 16.25cm

Mo = (26cm/16.25cm)

The magnificent given by objective lens, Mo = 1.6.

Optical devices: The camera

Lens camera is an instrument which produces an image of object on the screen using light. The basic physical principle of all camera is the same in spite of the variation in the design of cameras.

The optical system of the camera are very similar to that of the lantern projector but with the direction of light reversed.The converging lens forms a real image of the object to be photographed.(This image is diminished (smaller than the object and inverted)

The lens can be moved back and forward with the help of focusing any so that objects at different distances can be brought to the focus.A forced image is locate on the film or plate when the shuttled is open for a suitable amount of time as determined by the shutter speed.

Light enters the camera Box and makes a picture of the object on the film “( The film is sensitive to light)

The camera is equipped with a diagram or light entering the camera.It ensures that is incident centrally on the lens so that the distortion of the image formed is reduced

The aperture stop, which is the limiting diameter of the aperture thought which light enters the camera (given as fraction of focal length F of lens) is also called F Number.

This F Number; is the fraction of focal length of the lens given as focal length divide by lens diameter.

F number = Focal length, F/Lens diameter, d

FN = F/d

Where d = is lens diameter.

The Number Indicates the Number of times the focal length F of times the focal length F of the lens diameter ( or stop)
The smaller the F - Number for a given focal length the larger the lens diameter
The lens with a larger diameter has a greater light- gathering power or speed
This for such a lens the shutter allows light in the camera for a short interval of time.

Magnification of a lens camerais obtained as the ratio of the Image distance and the object distance.

But from the lens formula:

Thus M = v/U

I/U + I/V = I/F

I/V = I/F - I/U

(I/V) ^–I = ( U - F / FU)^-I

V = FU/ ( U - F)

Example 4

A lens camera is to be used to take a picture of a man 2m tall if the lens of the camera Has a focal length of 10cm, calculate the minimum size of the film frame required, given that the man is 20.1m from the camera.

Solution:

Magnification is given by:

M = f/ (u-f)

Where

F= 10cm U = 201/m / 2010cm

M = ( 10/2010 – 10)

M = 1/20 ....................................i

Let the size of the frame be h when the height of man is 2m.

Then

M = 1/200

But h1/h = 1/200

h1 = (1/200) 2

h1 = (1/200)2

h1 (2/200)

h1 = (1/100) m

h1 = 1cm or 10mm

The film frame should be at least 10mm square.

A simple lens camera

Problems involving the lens formula and the magnification formula

Physics Kenya

Thin Lenses

Physics Kenya Form 4 Topics

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