physics chapter 1- reflection of light at curved surfaces

Reflection of light by spherical mirrors

The first law of reflection tells us;

A light ray incident at an angle with the normal at the point of incidence

will get reflected making equal angle with the normal.

This law is true for all surfaces, be it a plane surface or a curved one.

The important words here are ‘the angle made with normal at the point of

incidence’. If for any surface one can decide the normal and find the incident

angle, it is possible to deduce the angle made by the reflected ray. It is very

easy to find a normal at any point on the plane surface but for a curved or

uneven surface it is not straight forward.

Activity 1

Finding the normal to a curved surface

Take a small piece of thin foam or rubber (like the sole of a slipper).

Put some pins along a straight line on the foam as shown in the figure -1a.

All these pins are perpendicular to the plane of foam. If the foam is

considered as a mirror, each pin would represent the normal at that point.

Any ray incident at the point where the pin makes contact with the surface

will reflect with the same angle as the incident ray made with the pinnormal.

Now bend the foam piece inwards as shown in figure-1b, what happens

to the pins?

They still represent the normal at various points, but you will notice

that all the pins tend to converge at a point (or intersect at a point).

If we bend the foam piece outwards, we will see that the pins seem to

move away from each other or in other words they diverge as shown in

figure-1c.

This gives us an idea of what is likely to happen with a spherical mirror.

A concave mirror will be like the rubber sole bent inwards (fig-1b) and the

convex mirror will be like the rubber sole bent out wards (fig-1c).For a concave mirror, like these pins in figure-1b, all normals will

converge towards a point. This point is called centre of curvature(C) of

the mirror.

This gives us a clue about how we can find normal to any point on a

spherical mirror. All that we have to do is to draw a

line from the point on the mirror to centre of the

sphere.

It is much easier to imagine this in a two

dimensional figure as shown in figure-2a. The

concave mirror is actually a part of a big sphere. In

order to find this centre point (centre of curvature)

we have to think of the centre of the sphere to which

the concave mirror belongs. The line drawn from

C to any point on the mirror gives the normal at

that point.

For the ray R, the incident angle is the angle it

makes with the normal shown as i and the reflected

angle is shown as r in figure-2b. We know by first

law of reflection i = r.

The mid point (Geometrical centre) of the mirror is called pole (P) of

the mirror. The horizontal line shown in the figures which passes through

the centre of curvature and pole is called principal axis of the mirror. The

distance between P and C is radius of curvature (R) of the mirror.

Try to construct different reflected rays for any array of rays that are

parallel to the principal axis as shown in figure 2(b). What is your

conclusion?

Verifying your drawing with experiment:

To verify this we must first find out some way of obtaining a beam of

parallel rays. How do we do that?

First we need to create a situation in which one gets parallel rays of

lightIn the figure-3 we have stuck two pins on a thermocole block. The

pins are parallel to each other. As we see in the figure, when a source of

light is kept very near, we see the shadows diverging

(from the base of the pins). As we move the source

away from the pins, the angle of divergence gets

reduced. If we move the source far away we will

get parallel shadows. But as we move the candle

away, the light intensity becomes low. That means

to get a beam of parallel rays the source should be

at a long distance and it must be of sufficient intensity.

Where do we find one such source?

Yes, we have one easily available source, you probably have guessed

it: The Sun.

Let us do an experiment with sun rays and a concave mirror.

Activity 2

Hold a concave mirror such that sunlight falls on it. Take a small paper

and slowly move it in front of the mirror and find out the point where you

get the smallest and brightest spot, which will be the image of the sun.

(See to it that your paper is small so that it does not obstruct the incoming

sun rays.)

The rays coming from the sun parallel to the

principal axis of concave mirror converge to a point

(see figure-4). This point is called Focus or focal

point (F) of the concave mirror. Measure the

distance of this spot from the pole of the mirror. This

distance is the focal length (f) of the mirror. The

radius of curvature will be twice this distance (R=2f).

Does this help you to verify the conclusions you arrived at, with your

drawing?

• What happens if you hold the paper at a distance shorter than the focal

length from the mirror and move it away?

• Does the image of the sun become smaller or bigger?

You will notice that the image of the sun first keeps on becoming small,

beyond the focal point it keeps on becoming enlarged.

Note: while drawing a ray diagram sometimes it is not clear which is

the reflecting side of the mirror. Hence we follow a convention of showing

lines on the non-reflecting side(coated side)Can you draw the same diagram for a convex

mirror?

See figure-5 The parallel rays appear to diverge

after reflection. If we extend the reflected rays

backwards they meet at ‘F’ i.e. focus of the convex

mirror

Lab Activity

Aim: Observing the types of images and measuring the object distance

and image distance from the mirror.

Material required: A candle, paper, concave mirror (known focal

length), V-stand, measuring tape or meter scale.

Procedure: Place the concave mirror on V-stand, arrange a candle and

meter scale as shown in figure-6.

Keep the candle at different distances from the mirror (10cm to 80cm)

along the axis and by moving the paper (screen) find the position where

you get the sharp image on paper. (Take care that flame is above the axis of

mirror, paper is below the axis)). Now we shall develop a technique to draw ray diagrams when

an object is placed anywhere on the axis of the mirror and validate the

above observations.

Here we will take at least two rays originating from the same point on

the object but with different direction, see how they get reflected from the

mirror and find out the point where they meet to form the image.

Let us take an example.

As shown in the figure-7, assume a concave mirror

and a candle placed at some distance along the axis of the

mirror.

The diagram shows two rays coming from the tip of

the flame (object). The reflected rays are constructed

based on the laws of reflection.They meet at point A. The

tip of the flame of the reflected image will be at the point

of intersection, A.

• Why only at point A?

If we hold the screen at any point before or beyond point A (for example

at point B), we see that the rays will meet the screen at different points.

Therefore, the image of the tip of the flame will be formed at different

points due to these rays. If we draw more rays emanating from the same tip

we will see that at point A they will meet but at point B they won’t. So, the

image of the tip of the flame will be sharp if we hold the screen at point A

and will become blurred (due to overlaping of multiple images) when we

move the paper slightly in any direction (forward or backward). Is this not

something that you observed during the previous experiment with sun rays?

However, it is not going to be easy to evaluate the angle of reflection

for any arbitrary ray, every time we will have to find the normal, measure

the incident angle and construct a ray with equal angle on the other side.

This would be a tedious task, can we find any other simpler method?

Yes, there are a few. Based on our discussion so far, we can identify

some appropriate rays which we can take as representative rays to find the

point ‘A’.

We have seen that all rays that are parallel to the

axis get reflected such that they pass through the focal

point of the mirror. So, for drawing any diagram the

most suitable ray to draw will be the one that comes

from the object and goes parallel to the axis of the

mirror. The reflected ray will be the line drawn from

the point of incidence on the mirror and passes through the focal point of the mirror. To make it more convenient we will always

take rays that come from the tip of the object. See the ray R1

in figure-8.

The converse situation of previous one is also true; that is, a ray that

passes through the focal point of the mirror will

travel parallel to the axis after reflection.

This gives us our second ray. This will be the

ray coming from the tip of the flame and going

through the focal point and falling on the mirror.

After reflection, this ray travels parallel to the axis.

So we draw the reflected ray as a line parallel to

the axis coming from the point where the incident

ray meets the mirror. See R2

in figure-9.

Using the rays R1

, R2

and finding the point where they intersect we

know the point where the image of the tip is formed.

There is one more ray which is convenient to draw.

We have seen earlier that any ray that is normal to the surface, on

reflection, will travel along the same path but in the opposite direction.

Which ray can such a one be for a spherical mirror?

We know that a line drawn from the centre of

curvature to the mirror is perpendicular to the tangent

at the point where the line meets the curve. So if we

draw a ray coming from the tip of the object going

through the centre of curvature to meet the mirror, it

will get reflected along the same line. This ray is shown

as R3

in the figure-10. In general, a ray travelling along

normal retraces its path.

Along with these three rays ‘the ray which comes from the object and

reaches the pole of the mirror’ is also useful in drawing ray diagrams. For

this ray, the principal axis is the normal.

If we have our object (candle) placed as shown in figure-11, we can

draw the the ray diagram to get the point of intersection A, of any two rays coming from the top of the object and point of intersection B, of any two rays coming from the bottom of the object. We notice that point B is exactly at the same distance from mirror as point A. Hence the image is vertical and inverted. • Where is the base of the candle expected to be in the image when the candle is placed on the axis of the mirror? Since any ray coming from any point on the axis and travelling along the axis will get reflected on the axis itself, we can conclude that the base of the image is going to be on the axis. Using the knowledge, that if the object is placed vertically on the axis, the image is going to be vertical, all that we need to do, is to draw a perpendicular from point A to the axis. The intersection point is the point where the base of the image of the candle is likely to be formed. See figure-12. Hence, as shown in the diagram the image will be inverted and diminished. Figure-12 is drawn for the case where the object is placed beyond the centre of curvature. Does this conclusion match with your observations? (Lab Activity) Draw similar diagrams for other cases and verify that they match with your observations. • During the experiment, did you get any positions where you could not get an image on the screen? Consider the case shown in the figure-13. The candle object (O) is placed at a distance less than the focal length of the mirror. The first ray (R1 ) will start from tip of the object and run parallel to axis to get reflected so as to pass through the focal point. This one is easy to draw. The second ray that we chose for earlier ray diagrams is the ray coming from the tip of the object and going through the focal point but it is not possible as such a ray will not meet the mirror. So we must use the third ray, a ray coming from the tip of the object and going through the centre of curvature. But that to does not seem to be possible. So we make a small change. Instead of drawing a ray from the candle tip to centre of curvature, we consider a ray that comes from the tip and goes in such a direction that it would go through the centre of curvature if extended backwards. This ray is normal to the surface and so will be reflected along the same line in opposite direction and will go through centre of curvature. We notice that the two reflected rays (figure-13) diverge and will not meet. While doing the experiments for a case such as this we were unable to find any place where we get a sharp image on the screen. This ray diagram tells us that since the reflected rays are diverging we will not get an image anywhere. So even if we had moved the screen much away from the mirror, we would not have found an image. In such situations, however, we do see an image when we look in the mirror. Is it possible to explain this image with a ray diagram? Remember what we did to find the image in a plane mirror. To decide the position of image we extended the reflected rays backwards till they meet. We will do the same here. When we look in the mirror we are looking at these diverging reflected rays. They appear to be coming from one point. We can get this point by extending the rays backwards as shown in figure14. The image does not really exist the way we see in other cases, but it is visible to us. As seen in the figure-14, the image will be erect and enlarged. Does this match with your observations? This image that we get by extending the rays backwards is called a virtual image. We cannot get this on a screen like a real image. The case in which the object is at the centre of curvature is another interesting situation. See figure-15. From the ray diagram (figure-15) we conclude that the image of the object will be formed at the same distance as the object and it will be inverted and of the same size. What is your observation? From the ray diagrams and the observations you may have noticed some peculiar properties of concave mirrors. They enlarge the image when the object is held close to the mirror (less than the focal length). Also, the image is erect. This property is used at many places and most commonly in shaving mirrors and mirrors used by dentists Another property is the way that it can converge the rays to its focal point. This is extensively used in many places. Look at the shape of TV dish antennas. If you look around you will see many curved surfaces but all surfaces are not concave, many of these are convex. Have you observed the rare view mirrors of a car? What type of surface do they have? Have you observed images formed on the rear and window glasses of a car? What type of surfaces are these? See figure -16. Can we draw ray diagrams for convex surfaces? Ray diagrams for convex mirrors One can draw ray diagrams for a convex mirror too. The ‘easy’ rays that we identified earlier can be used in this case with small modification. Here there are three rules which describe these rays. The procedure for drawing the diagram is similar and is not repeated here. Rule 1: A ray parallel to the axis, on meeting the convex mirror will get reflected so as to appear as if it is coming from the focal point. See figure-17. Rule 2: This is converse of Rule 1. A ray travelling in the direction of the focal point, after reflection, will become parallel to the axis. See figure-18. Rule 3: A ray travelling in the direction of the centre of curvature will, on reflection, travel in the opposite direction and appears to be coming from the centre of curvature. See figure-19.
Now let us use these rules to show the formation of images when the object is placed at different places infront of the convex mirror (see fig. - 20) AB is the object placed at any point on the principal axis infront of the convex mirror. Using Rule (1) and Rule (3), we get an erected, diminished, and virtual image between P and F on the back side of the mirror. This image can not be caught on screen and visible only in the mirror. Hence this is a virtual image. Verify this with an experiment. Using these rules, draw ray diagrams to show formation of images when an object is placed at different positions and note down your conclusions. Verify your results experimentally. You may get the image at a particular distance when you place the object at a certain distance. Do you find any relation between the object distance(u) and the image distance(v)?
Derivation of formula for curved mirrors Observe figure 21. The ray from the top (B) of the object AB is emanating parallel to principle axis and striking the mirror at X. This ray passes through F after reflection. Another ray starting from B, passes through centre of curvature (C) and strikes the mirror at Y. This ray returns back in the same direction after reflection. There two rays XBI and YBI are meeting at BI . So BI is the image of B. Hence the image of AB is AI BI . From the above figure - 21. ΔABC, ΔAI BI C are similar triangles AB AC AB AC = II I ..... (1) Draw a line XPI perpendicular to the principle axis. Similar ΔPI XF andΔAI BI Fare similar triangles PX PF AB AF = I I II I ..... (2)
From the figure - 21, we can say that PI X = AB AB P F AB AF = I II I ..... (3) from the equations (1), (3) we can write AC P F AC AF = I I I ..... (4) If the paraxial rays (rays which are travelling very near to principal axis) are considered, we can say that PI coincides with P Then PI F = PF AC PF AC AF = I I ..... (5) We can observe from fig. 21, that AC = PA − PC AI C = PC − PAI AI F = PAI − PF by substituting these in equation (5) PA PC PF PC PA PA PF − = − − I I ..... (6) we know that PA = u, PC = R = 2f, PAI = v, PF = f u 2f f 2f v v f − = − − (u − 2f) (v − f) = f(2f − v) uv − uf − 2vf + 2f 2 = 2f 2 − vf uv = 2f 2 − vf + uf + 2vf − 2f 2 uv = uf + vf ..... (7) Divide equation (7) with uvf uv uf vf uvf uvf uvf = + 1 11 f vu = + This is called mirror formula. While using this formula, we have to use sign convension in every situation.
Sign convention for the parameters related to the mirror equation 1. All distances should be measured from the pole. 2. The distances measured in the direction of incident light, to be taken positive and those measured in the direction opposite to incident light to be taken negative. 3. Height of object (ho ) and height of image (hi ) are positive if measured upwards from the axis and negative if measured downwards. Now let us understand magnification, i.e. the relation between the size of the object and the size of the image. Magnification (m) The image formed by a spherical mirror varies in size, here we discuss the variation in height only. Observe figure 22. A ray coming from OI is incident at pole with an angle of incidence θ, and get reflected with same angle θ. From fig. 22 we can say that ΔPOOI and ΔPIII are similar triangles. So III /OOI = PI/PO ................(1) according to sign convention PO = –u; PI = –v; OOI = h0 ; III = –hi Substuting the above values in equation (3). i o h v h u − − = − i o h v h u − ⇒ = ∴ Magnification m = i o h v h u − = We define the magnification, m = height of image(hi )/ height of object (h0 ) In all cases it can be shown that m = – image distance (v)/ object distance (u) Calculate the magnifications with the information you have in table2, for all the five cases. Example An object 4cm in size, is placed at 25cm infront of a concave mirror of focal length 15cm. At what distance from the mirror whould a screen be placed in order to obtain a sharp image?Find the nature and the size of the image.
Solution Accordint to sign convention: focal length (f) = –15cm object distance (u) = –25cm object height (ho ) = +4cm image distance (v) = ? image height (hi ) = ? Substitute the above values in the equation 1 f ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ = 1 1 u v ⎛⎞⎛⎞ ⎜⎟⎜⎟ + ⎝⎠⎝⎠ 1 11 15 v 25 ⎛ ⎞ ⎜ ⎟ = + ⎝ ⎠ − − B 111 v 25 15 ⎛ ⎞ = − ⎜ ⎟ ⎝ ⎠ 1 2 v 75 − = B v = –37.5cm So the screen should be placed at 37.5cm from the pole of the mirror. The image is real. magnification m = i o h v h u − = by substituting the above values i h ( 37.5) 4 ( 25) − = − − i (37.5 4) h (25) × =− hi = –6cm So, the image is inverted and enlarged. We have learnt the phenomenon of reflection of light by curved mirrors. Let us make use of it in our daily life. Making of solar cooker You might have heard the story of Archimedes burning ships using mirrors. Can we at least heat up a vessel using a mirror? Let us try: We have already learnt that a concave mirror focuses parallel sun rays at the focal point of the mirror. So with a small concave mirror we can heat up and burn paper as shown in the figure-23. (Try this with convex mirror also. What do you observe?)
In the same way make a big concave mirror to heat up a vessel. You might have observed the TV dish antenna. Make a wooden/ iron frame in the shape of TV dish. Cut acrylic mirror sheets in to 8 or 12 pieces in the shape of isosceles triangles with a height equal to the radius of your dish antenna. The bases of 8 or 12 triangles together make the circumference of the dish. Stick the triangle mirrors to the dish as shown in figure24. Your solar heater/cooker is ready. Arrange it so that concave part faces sun. Find its focal point and place a vessel at that point. The vessel gets heated enough to cook rice. In practical applications (like in car-headlights), concave mirrors are of parabolic shape (see figure 25)