Light: Sources of light, Magnification.

We shall be looking at Light wave in this module. Light is a visible form of energy radiated outward from a source. Light is the visible part of electromagnetic waves. Light is a typical example of transverse wave i.e it travels in a direction perpendicular to its vibration.

The energy associated with light is called luminous energy. Sources of light can be divided into two. They are;

  • Natural source
  • Artificial source
Natural source of light such as Sun and Star can be produce light on their own without mechanical means. Artificial source of light such as candle, electric light bulb, photographer flash bulb, can only produce light by mechanical or electrical means.

Object that can produce light are called Luminous objects. Examples are sun, star, candle, electric light bulb etc. Non Luminous Objects such as moon, book, person, stone etc do not produce light at all. They can only be seen when the rays of light from luminous object falls on them and reflect into the eyes.

TRANSMISSION OF LIGHT
Substances that allow light to pass through them are called Transparent substances. Examples are clean water, glass etc.
Substances that do not allow light to pass through them are called Opaque substances. Examples are called book, person, moon etc
Substances that allow light to pass through them but the source the light is coming from cannot be seen are called Translucent Substances. Examples are wax paper, laminated glass sheet etc.

RAY OF LIGHT
This is the path through which light travels. It is represented with a straight line and an arrow pointing to the direction of light. Two or more rays of light is called beam of light. It can be classified into three;

  • Convergent beam of light
  • Divergent beam of light
  • Parallel beam of light
Convergent beam of light refers to beam of light from different sources meeting at a point. Examples are rays of light falling on the retina, rays of light receive by camera falling on the film etc.

Divergent beam of light refers to beam of light from a source spreading into different directions. Examples are rays of light coming out from candle, electric bulb etc

Parallel beam of light travels in a parallel manner and they meet at infinity. searchlights etc
RECTILINEAR PROPAGATION OF LIGHT
This is the phenomenon that explains that light travels in a straight line. The three major effect of rectilinear propagation of light are Shadow, Eclipse and the pin-hole camera. Let's take a look at each of them.
SHADOW
This is the region where rays of light do not reach due to the obstruction by an opaque object. If a point source of light is use, there is a complete shadow that has smooth edges. If a large source of light is use, the shadow will have two region known as umbra and penumbra. Umbra is the region where the rays of do not reach while penumbra is the region where there is partial rays of light.
ECLIPSE

This occurs when the celestial bodies such as Sun, moon and the earth are in a straight line.
Eclipse of the sun also known as solar eclipse occur when the moon is between the sun and the earth. Eclipse of the moon also known as lunar eclipse occurs when the earth is between the sun and the moon. Annular eclipse occurs when the sun and the moon are in a position where the extreme rays of light intersect before reaching the earth.

PIN-HOLE CAMERA
It consist of a box with pinhole at one end and a wax paper which acts as a screen on the opposite end. If a luminous object is placed in front of the pinhole, it forms a real image that is inverted on the screen. The smaller the size of the hole, the smaller the image and it becomes sharp. A bigger hole will produce a bigger image but remain blurred.

Things we need to know about the pin hole camera
  • The distance between the object and the hole is called object distance (u)
  • The distance between the image and the hole is called image distance (v)
  • If we reduce the distance between the screen and the pinhole, the size of the image will decrease and the image will become bright because the light spreads over a small area.
  • If we increase the distance between the screen and the pinhole, the size of the image will increase and the image will become less bright because the light spreads over a large area.
  • If we increase the distance between the object and the pinhole, the size of the image will decrease.
  • If we decrease the distance between the object and the pinhole, the size of the image will increase.
Magnification
This is use to compare the size of the image and the object. If the image is bigger than the object, we say the image is magnified(i.e M>1). If the image is smaller than the object, we say the image is diminished (i.e M<1). If the image has the same size as the object, it means, M=1. Magnification has no units
These are the formulae for calculating magnification.

Let's take some questions on these
  • The blurred images produced by a pin-hole camera with a very small hole is attributed to (a) refraction (b) diffraction (c) polarization (d) reflection
The answer to the question is diffraction because diffraction is the spreading of waves after passing through an obstacle.
  • If the distance between the object and the pinhole camera is reduced by half, the size of the image of the object (a) remains the same (b) is halved (c) is quadrupled (d) is doubled
The answer is doubled
  • The pin-hole camera produces a less sharply defined image when the (a) screen is further from the pin-hole (b) pin-hole is larger (c) object is further from the pin-hole (d) illumination is less
The answer is (b)
  • An object is placed 10m from a pinhole camera of length 25cm. Calculate the linear magnification.
Solution
Length of camera= 25cm= 25/100= 0.25m, Object distance (u) = 10m

  • A man 1.5m tall is standing 3m in front of a pinhole camera whose distance between the hole and the screen is 0.1m. What is the height of the image of the man on the screen?
Solution
Object height= 1.5m, u=3m, v= 0.1m, image height=?

  • A pinhole camera is placed 300m in front of a building so that the image is formed on a screen 5cm from the pin hole. If the image is 2.5cm high, the height of the building will be
Solution
U=300m, v= 5cm=5/100=0.05m, image height=2.5cm=2.5/100=0.025m, object height=?




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