high school by ssl technologies physics ex-47 click the law of reflection when light strikes a...

High School

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Physics Ex-47

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THE LAW OF REFLECTION

When light strikes a surface and is reflected, it changes direction.The direction it takes depends upon the angle it strikes the surface.As illustrated below, a ray of light going towards a surface is calledan incident ray while a ray of light which is reflected away from asource is called a reflected ray.

The law of reflection says that the angle of the incident ray with thenormal equals the angle of the reflected ray (also with the normal).

Physics Ex-47

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By definition, the angle which the incident ray makes with the normalis called the angle of incidence while the angle which the reflectedray makes with the normal is called the angle of reflection.

Remember : The angle of incidence (i) is the angle formed by the incident ray and the normal (not the reflecting surface). The angle of reflection (r) is the angle between the reflected ray and the normal.

Note that the incident ray,the reflected ray,

and the normal are all on the same plane.

Physics Ex-47

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The law of reflection, therefore, simply states that when a ray oflight is reflected from a surface, it is reflected in such a directionthat the incident angle equals the reflected angle.

Using the law of reflection, we can determine the location of images formed by plane mirrors.

Physics Ex-47

Back of mirrorObject

An object is placed in frontof a plane mirror.

Find the location and characteristics of the image.

TASK

Rather than draw an actual

object, it is easier to draw

the object as an arrow.

This way, we can tell if the

image is inverted or not.

Objects consist of an infinite number of points.

And each point has an infinite number of rays radiating outwards.

Images also consist of an infinite number of points.

Each point on the object has a corresponding point on the image.

In locating an image, we take one point on the object and find its corresponding point (location) on the image object.

As shown above, we usually take the (extreme) top point and find its corresponding point on the image object.

Normal

Normal

Incident ray

Reflected ray

Extended ray

Incident ray

Reflected ray

Extended ray

We can now draw the image.

Since the two reflected rays diverge,

we must “extend” them so that they

will intersect.

There are two steps in finding the image of a plane mirror.NOTE

Draw an incident ray slightly upwards towards the mirror.Then draw its reflected ray in accordance with the law

of reflection.

Step-1

Draw an incident ray slightly downwards towards the mirror, then draw its reflected ray.

Step-2

Since we are not given the location of the object, we can select any arbitrary location.

Such asright here!

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Where the two reflected rays meet is the image point of the object point.

Physics Ex-47

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Virtual images are formed by extended rays.

Real images are formed by reflected rays.

Physics Ex-47

Physics Ex-47

Eye

Mirror

Given a plane mirror and the position of an observer’s eye, determine the field of vision.

TASK

Field of vision

(Any point within this area will be reflectedto the observer’s eye as an image point.)

There are two steps in finding the field of vision.NOTE

Draw a ray from the eye to one end of the mirror.This is the reflected ray. Then, in accordance with

the law of reflection, draw the incident ray.

Step-1We can now draw the field of vision.

It is the area bounded by the mirror

and the two incident rays.

Refl

ecte

d ra

y

Incident ray

Reflected ray Incident ray

angle of incidence = angle of reflection

angle of incidence = angle of reflection

Draw a ray from the eye to the other end of the mirror.This is the reflected ray. Then, in accordance with

the law of reflection, draw the incident ray.

Step-2

NormalNormal

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Physics Ex-47

Eye

Mirror

Given a plane mirror, a pointobject (P) and the position of an observer’s eye, determine whether or not the observer

can see object point P.

TASK

P

P'

To determine whether or not the observer can see the given object,we first locate the image of the object.

Next, we draw a line from the image point to the observer (eye).

If the line intersects the mirror, the observer can see the image.If the line does not intersect the mirror, then the observer cannotsee the image.

NOTE

Object point

To find the location of the image point, we start by drawing a lineperpendicular to the mirror and extend it beyond the mirror.

Perpendicular to mirrorNext, since the image distance equals the object distance,we mark off the image point, P‘, along the extended line.Object

distance

Imagedistance

Image point

Finally, we draw a straight line from theobserver’s eye to the image point.

If this line intersects the mirror,the observer can see the image.

If this line does not intersect the mirror,the observer cannot see the image.

Look! The line does not intersect the mirror.

Conclusion: observer cannot see the image. ANSWER

The observer cannot see the image.

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Object distance = Image distance

Physics Ex-47

Eye

Mirror

Given a plane mirror, a pointobject (P) and the position of an observer’s eye, determine whether or not the observer

can see object P.

TASK

P

P'

To determine whether or not the observer can see the given object,we first locate the image of the object.

Next, we draw a line from the image point to the observer’s eye.

If the line intersects the mirror, the observer can see the image.

If the line does not intersect the mirror, then the observer cannotsee the image.

NOTE

Object point

Perpendicular to mirror

Image point

To find the location of the image point, we start by drawing a lineperpendicular to the mirror and extend it beyond the mirror.

Objectdistance

Imagedistance

We now draw a straight line from theobserver’s eye to the image point.

If this line intersects the mirror,the observer can see the image.

If this line does not intersect the mirror,the observer cannot see the image.

Look! The line does not intersect the mirror.

Thus, the observer cannot see the image.

ANSWER

The observer cannot see the image.

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Object distance = Image distance

Physics Ex-47Question-1

State the Law of Reflection.

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The angle of incidence equals the angle of reflection.

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The diagram below represents an object in front of a plane mirror.

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Note : Diagram not drawn to scale.

a) Draw the image.

b) How high is the image? 4 cm

c) How far is the image from the object?

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20 cm

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Object

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Explain why there is only a lateral (left-right) reversal whenwe look at ourselves in a plane mirror.

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Because when we turn towards a mirror, we do so by turning aboutthe Y-axis (left-right) and not about the X-axis (top-bottom).

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The diagram below illustrates the image of an object producedby a plane mirror. Label the incident ray, the reflected ray andthe extended ray.

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Reflected ray

Incident rayExtended ray


Explain how real images are formed and how virtual imagesare formed.

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Real images are formed by the intersection of reflected rays,

virtual images are formed by the intersection of extended rays.

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NOTE

As long as there is at least one extended ray,the image is said to be virtual.


Stefania is 1.5 m tall. Prove that the shortest mirror necessaryfor Stefania to see her full height is 75 cm (half her height).

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Divide Stefania’s height into two parts, from her eyes to the top ofher head and from her eyes to the bottom of her feet.

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Thus, in order for a person to see their full height in a plane mirror,the minimum mirror required is a mirror half their height.

Half Stefania’s height


A beam of light is reflected from a plane mirror such that theangle between the incident ray and reflected ray is 50o.

Draw the beam and calculate the angle of incidence? .

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Two mirrors, M1 and M2, are at 90o to each other. As illustrated,a beam of light strikes mirror M1 with an angle of incidence of 60o and is reflected by mirror M2. Complete the diagram and find theangle of incidence of the ray reflected by mirror M2.

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Angle of incidence of mirror M2


Two plane mirrors, M1 and M2, are at 60o to each other as illustratedin the diagram below. A beam of light strikes mirror M1 with an angleof incidence of 40o. Complete the diagram and determine the angleof reflection of the beam reflected from mirror M2 ?

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Angle of reflection of mirror M2


Two mirrors are parallel to each other as illustrated in the diagrambelow. A beam of light strikes the beginning of one mirror at anangle of incidence of 35o. Complete the diagram and find thenumber of times the beam is reflected before it emerges fromthe two mirrors? .

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Physics Ex-247

A) 1

B) 2

C) 3

D) 4

Question-11

What is the angle of reflection for the ray diagram illustrated below?

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REMINDERAll angles are measured

from the “normal”.

Angle of incidence

Angle of reflection


Stefania wants to install a rear view mirror on her bike. She has achoice of choosing a plane or a convex mirror as illustrated below.

Which mirror should Stefania choose in order to obtain the maximum filed of view?

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Observer

Observer

Answer:

Stefania should choose theconvex mirror in order to

obtain the maximum fieldof view.


The following set up consists of a light source, an obstacle,six different locations (labeled 1 to 6) and a target.

At which one of the locations can you place a plane mirrorso that the light source is reflected to the target?

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To determine the correct location,we draw two lines from each point.

One line to the light source andanother line to the target. If bothlines are unobstructed, that is thepoint where we place the mirror.

Note that only at location 5 canwe draw unobstructed lines bothto the source and to the target.

Thus, we must place the mirrorat location 5.

No

No

No

No

Yes

No

We now divide the angle between the two lines at location 5 with a bisector. This bisector will be the

normal to the mirror.

Finally, we draw the mirror perpendicular to the normal (bisector).

Mirror

If required, using a protractor, we can measure the angle of incidence or

the angle of reflection.


The set up below consists of an object and its shadow cast ona screen by a point light source.

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Determine the position of the light source.

A) Point-A

B) Point-B

C) Point-C

D) Point-D

Draw a line from the top of the shadowto the top of the object and then extend

the line to the light source.

This problem illustrates the fact that light travels in a straight line, technically known

as rectilinear propagation .

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