# Application of Refraction & Reflection

Tags: mirror

## Laws of Refraction

(i)   The incident ray, the refracted ray and the normal to the refracting surface at the point of incidence all lie in the same plane.

(ii)  The ratio of the sines of the angle of incidence (i) and that of the angle of refraction (r) is a constant quantity for two given media.

 This law is known as Snell’s law.             The value of the constant is equal to the relative refractive index µ21 of the second medium with respect to the first medium.             The refractive index of a substance relative to a vacuum (a ray passes through the vacuum to the given medium) is called the absolute refractive index of the substance.

The relative refractive index for two media is equal to the ratio of their absolute refractive indices.

where µ2 and µ­1 are the absolute refractive indices of the second and first media.

The absolute refractive index of a medium is defined as the ratio of the speed of light in vacuum to the speed of light in the medium.

 The relative refractive index 1m2 is given by

Illustration-9

Refractive index of glass with respect to water is 1.125. Find the absolute refractive index of water, if the absolute refractive index of glass is 1.5.

Solution

Illustration-10

The velocity of light in air is 3 × 108 m/s. Find the velocity of light in glass of refractive index 1.5. If the wavelength of yellow light in air is 6000Å, find its wavelength in the glass.

Solution

Let c1 and c2 be the velocities of light in air and glass respectively.

Using the definition of refractive index.

According to Snell’s law

Thus, sin θ12 ⇒            θ12

If light is incident normal to a boundary (i.e. i = 0), then, it passes undeviated from the boundary as shown in the fig.(7 a).

 If the refractive indices of the two media are equal as shown in fig.(7 b), then also the light ray is not refracted and the boundary between the two media is not visible. This is why a transparent solid is invisible in a liquid of same refractive index.

Basic Problem in Optics

To find the position and nature of the final image formed by certain optical systems for a given object. The optical system may be just a Mirror, or a lens or a combination of several reflecting and refracting surfaces.

Basic Strategy

First of all, we identify the sequence in which the reflection and refraction are taking place. The several events of reflection or refraction can be named as event 1, event 2 and so on following the sequence in which they occur. Now, the image of event 1 would be object for event 2, image of event 2 will be object of event 3 and so on. This way one can proceed to find the final image.

Cartesian Sign Convention

 (i)     All distances are measured from the pole. (ii)   Distances measured in the direction of incident rays are taken as positive. (iii) Distances measured in a direction opposite to that of the incident rays are taken as negative. (iv) Distances above the principal axis are taken as positive. (v)   Distances below the principal axis are taken as negative. (vi) Angles measured from the normal, in anti-clockwise direction are positive, while in clockwise directions are negative.

REFLECTION FROM A PLANE-SURFACE

When a real object is placed in front of a plane mirror (as shown in the figure), the image is always erect, virtual and of same size as the object. It is at same distance behind the mirror as the object is infront of it.

An optical image is a point where reflected or refracted rays of light either intersect or appear to intersect. Thus, the image of a infinite object is an assembly of image points corresponding to various points of the object. If the rays after refraction or reflection actually converge at a point, the image is said to be real and it can be obtained on a screen. However, if the rays do not actually converge but appear to do so the image is said to be virtual. A virtual image cannot be obtained on a screen.

The image formed by a plane mirror suffers lateral-inversion i.e., in the image formed by a plane mirror, left is turned into right and vice-versa with respect to object as shown in figure. However, mirror does not turn upside and down and vice-versa. The reason for this is that mirror actually reverses forward and back in three-dimensions (and not left into right) – i.e., only z – direction is reversed resulting in the change of left into right or vice-versa. Actually a plane mirror changes right-handed co-ordinate system (or screw) to left handed.

 Keeping the incident ray fixed, if the mirror is rotated through an angle θ, about an axis in the plane of mirror then the reflected ray rotates through an angle 2θ, as shown in the fig.(14).
 Illustration–11             A ray of light is incident at an angle of 30o with the horizontal. At what angle with horizontal must a plane mirror be placed in its path so that it becomes vertically upwards after reflection? Solution
• REFLECTION FROM CURVED SURFACES

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.(16).             Applying geometry, we get                         γ = a + i             and      β = r + γ             Since   i = r (law of reflection)                     -α + β = 2γ             Applying paraxial ray approximation

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.

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

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

1. 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.
1. 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
1. 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.

(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.(24 b).

1. Concave mirror behaves as convex lens (both convergent) while convex mirror behaves as concave lens (both divergent). This is shown in fig.(25).
1. 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 1   Position and nature of image formed for a given position of object.

(a) For concave mirror

### Illustration-12

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.

### Illustration–13

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

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Application of Refraction & Reflection

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