# Reflection from Curved Surfaces

## Optics of Class 12

### Important Terminology

(i)Centre of curvature : It is the centre-of the sphere of which the glass used in mirror/lens is a part.

(ii)Radius of curvature : It is the radius of the sphere of which the glass is a part used in mirror or lens.

(iii)Pole : It is the geometrical centre of the spherical reflecting surface.

(iv)Principal axis (for a spherical mirror) : It is the straight line joining the centre of curvature to the pole.

(v)Focus : When a narrow beam of rays of light, parallel to the principal axis and close to it, is incident on the surface of a mirror (lens), the reflected (refracted) beam is found to converge to or appears to diverge from a point on the principal axis. This point is the focus.

(vi)Focal length (for a mirror) : It is the distance between pole and the principal focus.

(vii)Real image : If reflected (or refracted) rays converge to a point (i.e. intersect there), the point is a real image.

(viii)Virtual image : If reflected (or refracted) rays appear to diverge from a point, the point is a virtual image.

(ix)Real object : If the incident rays diverge from a point, the point is a real object.

(x)Virtual object : If incident rays converge and appear to intersect at a point behind the mirror (or lens), the point is a virtual object.

### Mirror Formula

 Consider a point object O placed on the principal axis of a concave mirror as shown in the fig.(15.16). Applying geometry, we get γ = α + i andβ = r + γ Since i = r (law of reflection) ∴α + β = 2γ Applying paraxial ray approximation

tanα ≈ α = AP/x; tanβ ≈ β = AP/y; tanγ = γ = AP/R

Thus, AP/x + AP/y = 2(AP/R)

or1/x + 1/y = 2/R = 1/f(since f = R/2)(15.5)

A similar formula can be obtained for the convex mirror as shown in the figure.

 Using geometry, i = α + γ β = r + γ Since i = r, therefore, -α + β = 2γ Applying paraxial ray approximation -1/x + 1/y = 2/R = 1/f (15.6)

Using cartesian convention the mirror formula can be generalised as

1/v + 1/u = 1/f (15.7)

In the case of concave mirror: u = -x, v = -y, f = -f

1/-x + 1/-y = 1/-f

or 1/x + 1/y = 1/f

Thus, we obtain the result shown by equation (15.5).

In the case of convex mirror: u = -x, v = +y and f = +f

1/-x + 1/y = 1/f

Thus, we get the result as shown by equation (15.6).

### Rules for Image Formation

These are based on laws of reflection i.e. i = r and are used to find details of image formed by a mirror. Usually the following rules are followed

1.A ray passing parallel to principal axis after reflection from the mirror passes or appear to pass through its focus.

2.A ray initially passing through or directed towards focus after reflection from the mirror becomes parallel to the principal axis.

3.A ray initially passing through or directed towards centre of curvature, after reflection from the mirror retraces its path.

4.Incident and reflected rays at the pole of a mirror are symmetrical about the principal axis. (As for pole principle axis acts as normal and from laws of reflection i = r).

So by observing the size of erect image in a mirror we can decide the nature of the mirror i.e., whether it is convex, concave or plane.

### Important Points

1.For real extended objects, if the image formed by a single mirror is erect it is always virtual and in this situation if the size of the image is-

2.For real extended object, if the image formed by a single mirror is inverted, it is always real (i.e., m is -ve) and the mirror is concave. In this situation if the size of image is−

3.Every part of a mirror forms complete image. If some portion of a mirror is obstructed (say covered with black paper), then complete image will be formed but intensity will be reduced.

4.If an object is moved at constant speed towards a concave mirror from infinity to focus, the image will move (slower in the beginning and faster later on) away from the mirror. This is because, in the time object moves from ∞ to C the image will move from F to C and when object moves from C to F the image will move from C to ∞. At C the speed of object and image will be equal.

5.As focal-length of a spherical mirror f (= R/2) depends only on the radius of mirror and is independent of wavelength of light and refractive index of medium so the focal length of a spherical mirror in air or water and for red or blue light is same. This is also why the image formed by mirrors do not show chromatic aberration.

6.In case of spherical mirror if R → ∞ (i.e., it becomes plane), f = R/2 = ∞, the mirror formula

1/v + 1/u = 1/f reduces to 1/v + 1/u = 0i.e.,v = −u

i.e., image is at same distance behind the mirror as the object is infront of it.

This in turn verifies the correctness of mirror formula.

7.In case of spherical mirrors if object distance (x1) and image distance (x2) are measured from focus instead of pole, u = (f + x1) and v = (f + x2) so the mirror formula

1/v + 1/u = 1/f reduces to 1(f + x2) + 1/(f + x1) = 1/f

which on simplification gives

x1x2 = f 2(15.9)

This result is called 'Newton's formula'.

8.In case of spherical mirrors if we plot a graph between −

(a)(1/u) and (1/v) the graph will be a straight line with intercept (1/f) with each axis as 1/v + 1/u = 1/f becomes y + x = c with c = 1/f. This is shown in fig.(15.24 a).

(b)u and v the graph will be a hyperbola as for u = f, v = ∞ and for u = ∞, v = f. A line u = v will cut this hyperbola at (2f, 2f). This all is shown in fig.(15.24 b).

9.Concave mirror behaves as convex lens (both convergent) while convex mirror behaves as concave lens (both divergent). This is shown in fig.(15.25).

10.As convex mirror gives erect, virtual and diminished image, field of view is increased. This is why it is used as rear-view mirror in vehicles. Concave mirrors give enlarged erect and virtual image (if object is between F and P) so are used by dentists for examining teeth. Further due to their converging property concave mirrors are also used as reflectors in automobiles head lights and search lights and by ENT surgeons in opthelmoscope.

Table 15.1 Position and nature of image formed for a given position of object.

### (a) For concave mirror

 S.No. Position of Object Ray-Diagram Details of Image 1. At infinity Real inverted, very small (m << -1), at F 2. Between ∞ and C Real, inverted, diminished (m < -1), between F and C. 3. At C Real, inverted, equal (m = -1), at C 4. Between F and C Real, inverted, very large (m >>-1), between F and C 5. At F Real, inverted, very large (m>>-1), at infinity. 6. Between F and P Virtual, erect, enlarged (m > +1) behind the mirror)

### (b)For convex mirror

 S.No. Position of Object Ray-Diagram Details of Image 1. At infinity Virtual, erect, very small (m << +1) at F 2. In front of mirror (not too far from the mirror) Virtual, erect diminished (m < +1) between P and F

Example 15.4

A rod of length 20 cm is placed along the optical axis of a concave mirror of focal length 30 cm. One end of the rod is at the centre of curvature and the other end lies between F and C. Calculate the linear magnification of the rod

Solution

 The image of the end a of the rod is formed at its position. The position of the image of end b can be obtained using mirror formula. 1/v + 1/u = 1/f u = -40 cm, f = -30 cm ∴ 1/v + 1/-40 = 1/-30 Thus,v = -120 cm Length of the image (rod) is li = 120 – 60 = 60 cm ∴magnification m =-Ii/Io = -60/-20 = +3

Example 15.5

Where should an object be placed in front of a concave mirror of focal length 30 cm so the image size is 5 times the object size?

Solution:

Mirror formula 1/v + 1/u = 1/f

or u/v + 1 = u/f

Since m = -v/u therefore, m = f/f-u

Here f = -30 cm and m = ±5 because the image can be real or virtual.

For real image: m = -5

∴-5 = -30/-30 - u

oru = -36 cm

For virtual image: m = +5

∴5 = - 30/-30 - u

oru = 024 cm

Hence, the object must be placed at 24 cm or 36 cm in front of the concave mirror.

Example 15.6

An object of height 2.5 cm is placed at a 1.5 f from a concave mirror, where f is the magnitude of the focal length of the mirror. The height of the object is perpendicular to the principal axis. Find the height of the image. Is the image erect or inverted?

 Solution Focal length of a concave mirror is negative. Nowu = -1.5 f Applying mirror formula orv = -3 f Lateral magnification is given by m = -v/u = -(-3f/-1.5f) = -2

since = hi /ho = -2 m =∴ hi = -2ho = -5 cm

The image is 5 cm long. The minus sign shows that it is inverted.