Reflection, Refraction and Total Internal Reflection

Ray optics as a limiting case of wave optics

In my optics (also called geometrical optics) the light is supposed to be propagating in straight lines, called the light rays, which are supposed to be formed of corpuscles. In wave optics light is supposed to be propagating in the form of waves.

The phenomena of interference, diffraction and polarization could only be explained by wave theory. The diffraction of light is the bending of light round the edges of the obstacle, due to which sharp images of the objects may not be seen.

The phenomenon of diffraction is more noticeable if the size of the object is comparable to the wavelength of light. According to wave theory the path of light may only be rectilinear approximately and not exactly. When the size of aperture becomes much greater than the wavelength of light, the light follows the straight line path. Therefore it may be concluded that the ray optics is a limiting case of wave optics.

Reflection

Laws of Reflection

The regular reflection follows the two laws:

1. The incident ray, the reflected ray and normal to surface at the point of incidence all lie in the same plane.

2. The angle of incidence (i) is equal to the angle of reflection (r’).

(i) Formation of Image by the plane mirror: The formation of image of a point object O by a plane mirror is represented in figure below.

The image formed ‘I’ has the following characteristics.

Formation of Image by the plane mirror

(a) The size of image is equal to the size of object.

(b) The separation of image from mirror formed behind the mirror is equal to the separation of object from the mirror i.e. OM = MI.

Number of images in inclined mirrors:

Let $\theta$ be the angle between two plane minors:

(i) If the object is placed asymmetrically between mirrors, no. of images $n = 360 / \theta$

(ii) If the object is placed symmetrically between mirrors and the value of $360 / \theta$ is even , then $n = ( 360 / \theta ) -1$ .

(1 is subtracted because two images coincide)

Refraction

When a ray of light falls on the boundary separating the two media, there is a change in direction of ray. This phenomenon is called refraction.

Laws of Refraction:

(i)The incident ray, the refracted ray and normal to the surface separating the two media all lie in the same plane.

(ii) Snell’s Law: For two media, the ratio of sine of angle of incidence to the sine of the angle of refraction is constant for a beam of particular wavelength I.e.

$\dfrac{sin \, i}{sin \, r}= constant = \dfrac{ \mu _2}{ \mu _1} = 1 \mu _2$

Where $\mu_1$ and $\mu_2$ are absolute refractive indices of ‘I’ and ‘II’ media respectively and $1 \mu_2$ is the refractive index of second medium with respect to ‘I‘ medium.

Snell’s Law

As light flows reversible path, we have:

$\dfrac{sin \, r}{sin \, i} = 2 \mu_1 \cdots Equation \, \, 1$

Multiplying equation 1 and 2 we get:

$2 \mu_1 \times 1 \mu_2 \, \, \, \, or \, \, 1 \mu _1 = \dfrac{1}{1 \mu_2} \cdots Equation \, \, 3$

Also the frequency of light remains unchanged when passing from one medium to the other.

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

I.e.

$\mu = \dfrac{Speed \, \, of \, \, light \, \, in \, \, vaccum}{Speed \, \, of \, \, light \, \, in \, \, Medium} = \dfrac{c}{v}$

$= \dfrac{v \lambda_{air}}{v \lambda_{medium}} = \dfrac{\lambda_{air}}{\lambda_{medium}} \cdots Equation \, \, 4$

$\lambda_{air} \, \, and \, \, \lambda_{medium}$ being wavelength of light in air and medium respectively.

$\therefore \dfrac{sin \, i}{sin \, r} = \dfrac{\mu_2}{\mu_1} = \dfrac{c/V_2}{c/v_1} = \dfrac{v_1}{v_2} = \dfrac{\lambda_1}{\lambda_2} \cdots Equation \, \, 5$

Formation of image by Refraction

According to Snell’s law if $\mu_2 > \mu_1 , I > r$ . That is if a ray of light enters from rarer medium to a denser medium, it is deviated towards the normal and if $\mu_2 < \mu_1 , i< r$ that is if the ray of light enters from denser to a rarer medium it is deviated away from the normal.

Accordingly if the ray of light starting from objects ‘O’ in denser medium travels along OP, it is deviated away from the normal along PQ. The ray PQ appears to come from ‘I’.

Thus ‘I’ is the virtual image of ‘O’. It can be shown that:

$Formation of image by Refraction$

Formation of image by Refraction

$\mu = \dfrac{Real \, \, depth ( OM )}{Apparent \, \, depth ( M I )} = \dfrac{t}{t-x} \cdots Equation \, \, 6$

Where ‘x’ is displacement or apparent shift.

$\therefore The \, \, apparent \, \, shift, \, \, x = ( 1 - \dfrac{1}{\mu} t ) \cdots Equation \, \, 7$

Refraction through a number of media

Now let us consider the refraction of light ray through a series of media as shown in figure. The ray AB is incident on air-water interface at an angle ‘I’. The ray is deviated in water along BC towards the normal.

Then it falls on water-glass interface and is again deviated towards normal along CD. If the last medium is again air, the ray emerges parallel to the incident ray. Let $r_1 and r_2$ be angles of refraction in water and glass respectively.

Then from Snell’s law,

$Refraction through a number of media$

Refraction through a number of media

$\dfrac{sin i}{sin r} = \dfrac{\mu_w}{\mu_a} = _a \mu_w \cdots Equation \, \, i$

$\dfrac{sin \, r_1}{sin \, r_2} = \dfrac{\mu_g}{\mu_w} = _w \mu_g \cdots Equation \, \, ii$

$\dfrac{sin \, r_2}{sin \, i} = \dfrac{\mu_a}{\mu_g} = _g \mu_a \cdots Equation \, \, iii$

Where,

$\mu_a$ = refractive index of air = 1

$\mu_w$ = refractive index of water

$\mu_g$ = refractive index of glass

Multiplying equation (i), (ii) and (iii), we get:

$_a \mu_w \times _g \mu_a = 1$

$_w \mu_g = \dfrac{1}{ _a \mu_w \times _g \mu_a} = \dfrac{_a \mu_g}{a \mu_w} \cdots Equation \, \, 8$

Lateral shift on passing through a glass slab: Consider refractive of a ray AO incident on the slab at an angle of incidence ‘I’ through the glass slab EFGH.

After refraction the ray emerges parallel to the incident ray.

$Refraction through a number of media$

Refraction through a number of media

Let PQ be perpendicular dropped from P on incident ray produced.

The lateral displacement caused by plate,

X = PQ = OP sin (I – r)

$\dfrac{OM}{cos r} sin (i-r) = \dfrac{t sin (i-r)}{cos r}$

(iii) If ‘I’ is very small, r is also very small, then:

$sin I \rightarrow I, \, \,sin r \rightarrow r \, \, and \, \, cos r \rightarrow 1$

So that $\dfrac{sin \, i}{sin \, r} = \mu$ takes the form $\dfrac{i}{r} = \mu$ .

Therefore, the expression for lateral displacement takes the form:

$x = \dfrac{t(i-r)}{1} = t \, I ( 1 - \dfrac{r}{i} )$

$= ( 1 - \dfrac{1}{\mu} t \, i )$

Critical Angle: Total Internal Reflection

The angle of incidence in denser medium for which the angle of refraction in rarer medium is 90° is called the critical angle (C).

If $\mu_r$ and $\mu_d$ are refractive indices for rarer and denser media then,

$\therefore \dfrac{sin \, i}{sin \, r} = \dfrac{\mu_2}{\mu_1} \, gives \\[3mm] \dfrac{sin \, C}{sin 90^o} = \dfrac{\mu _r}{\mu _d} = _d \mu_r$

$\therefore sin \, C = _d \mu_r = \dfrac{1}{_r \mu_d} = \dfrac{1}{\mu}$

Where $_r \mu_d = \mu$ it is the refractive index of denser medium with respect to rarer medium. When angle of incidence of the ray incident on rarer medium from denser medium is greater than the critical angle, the incident ray does not refract into rarer medium but is reflected back into denser medium. This phenomenon is called total internal reflection.

The conditions for total internal reflection are:

(i) The ray must travel from denser to rarer medium.

(ii) The angle of incidence i > critical angle C.

The critical angle for water-air, glass-air and diamond air interfaces is 49°, 42° and 24° respectively.

A fish or diver in water at depth h sees the whole outside world in horizontal circle of radius,

$r = h tan C = \dfrac{h}{ \sqrt{ ( \mu ^2 - 1}}$

$\mu$ being refractive index of water.

Optical fibre

Optical fibre is a device based on total internal reflection by which signals may be transferred from one location to another. It is a thin pipe of plastic or specially coated glass in which light enters at one end and leaves at other end suffering a number of total internal reflections with little loss of energy.

Optical fibre

The fibre works even if it is bent or twisted. For total internal reflection at the wall of fibre, the angle of incidence i > C, where sin C $= 1 / _a n_f , _a n_f$ being refractive index of fibre with respect to air. The thickness of fibre is of the order of human hair = $50 \mu = 50 \times 10^{-6} m$ .

A bundle of optical fibres can be put to several uses :

(1) It can be used as a ‘light pipe’ in medical and optical examination.

(2) It can transmit a laser or any other light beam.

(3) They are being used in telephone and other transmitting cables.

Sign Conventions

The sign conventions of coordinate geometry will be used, taking pole of mirror as origin. Accordingly the focal length of concave mirror is negative and that of convex mirror is positive.

The distance of object placed in front of mirror on the left (u) is (Negative X-axis) negative and the distance of image from mirror (v) is negative for real image and positive for virtual image.

Curved Mirrors and Mirror formulae

There are two types of spherical (curved) mirrors:

(i) Convex: Convex mirrors forms only virtual images of a real objects.

(ii) Concave: Concave mirrors may form real and virtual images or real objects.

Mirror formulae for all spherical mirrors are:

$\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$

$f = \dfrac{R}{2}$ and magnification $\dfrac{I}{O} = - \dfrac{v}{u} = - \dfrac{f}{u-f}$

For a convex mirror, f is positive and for a concave mirror f is negative.

If the object is to the left of the mirror u is negative and v is positive if image is on the right and negative if image is on the left of mirror.

Image formation by concave mirror

Image formation by convex mirror

It is observed that a spherical mirror of large aperture does not give a sharp image because the marginal rays (outer rays) are focused at a relatively smaller distance from pole P. This defect in image is called spherical aberration.

This is reduced by taking spherical mirror of small diameter as compared to its length focal or it is completely eliminated by taking parabolic mirrors.

Lens and Lens Formulae

There are two types of lenses:

(i) Convex (or converging) lens

(ii) Concave (or diverging) lens

Lens Maker’s Formula

If $R_1 and R_3$ are the radii of curvature of first and second refracting surfaces of a thin lens of focal length f, then lens-makers formula is

Lens Maker's Formula

$\dfrac{1}{f} = ( 1 \mu_2 -1 ) ( \dfrac{1}{R_1} - \dfrac{1}{R_1} )$

$= ( \mu - 1 ) ( \dfrac{1}{R_1 - \dfrac{1}{R_1} } )$

Where $1 \mu_2 = \mu$ is refractive index of material of lens with respect to surrounding medium.

Thin lens formula is:

$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$

Magnification produced by a lens:

$m = \dfrac{I}{O} = \dfrac{v}{u}$

Where ‘I’ is size of image and O, is size of object.

If a lens (refractive index $\mu_2$ ) separates two media of refractive indices $\mu_1 \, \, and \, \, \mu_3$ then its total length ‘f’ is:

$\dfrac{\mu_3}{f} = \dfrac{\mu_2 - \mu_1}{R_1} - \dfrac{\mu_3 - \mu_2}{R_2}$

Thin lens formula

Power of lens: The power, of a lens is its ability to deviate the rays towards axis and is given by:

$P = \dfrac{1}{f ( in \, \, meters ) } Diopters = \dfrac{100}{f ( in \, \, cm ) } Diopeters$

Lens immersed in a liquid: If a lens refractive index $\mu_g$ is immersed in a liquid of refractive index $\mu_l$ then its focal length ( $f_I$ ) in liquid, is given by:

$\dfrac{1}{f_1} = ( _1 \mu_g -1 ) ( \dfrac{1}{R_1} - \dfrac{1}{R_2} )$

Where,

$_1 \mu_g = \dfrac{\mu_g}{\mu_i}$

(i) If $f_a$ is the local length of lense in air, then,

$f_l = \dfrac{n_g -1}{ \dfrac{n_g}{n_l} -1} \times f_a$

Now there arise three cases:

(i) If $\mu_g > \mu_l \, \, then f_l \, \, and \, \, f_a$ are of same sign but $f_l > f_g$ .

That is the nature of lens remains unchanged, but its focal length increases and hence power of lens decrease. In other words the convergent lens becomes less convergent and divergent lens becomes less divergent.

(ii) id $\mu_g = \mu_l , \, \, then \, \, f_l \rightarrow \infty$ . That us the lens behaves as a glass plate.

(iii) if $\mu_g < \mu_l$ the $f_l \, \, and \, \, f_q$ have opposite signs.

That is the nature of lens changes. A convergent lens becomes divergent and vice versa.

Newton’s formula

If the distances of object and image are not measured from optical center, but from first and second principal foci respectively, then,

Newton’s formula states $f_1 f_2 = x_1 x_2$

Newton’s formula

Where $x_1 = F_1 O$ = distance of object from I principal focus $F_1$ .

$x_2 = F_2 I$ = distance of image from II principal focus $F_2$ .

If medium on either side on lens is same, then $f_2 = -f_1 = f$

Therefore, newton’s formula takes the form, $x_1 x_2 = -f^2$ .

Lenses in contact: If two or more lenses of focal lenses of focal lengths $f_1 , f_2 \cdots$ are placed in contact, then their equivalent focal length F is given by:

$\dfrac{1}{F} = \dfrac{1}{f_1} + \dfrac{1}{f_2} + \cdots = \sum \dfrac{1}{f}$

The power of communication $P = p_1 + p_2 + \cdots = \sum P$

Displacement method

For real image the distance between object and screen must be greater than or equal to 4f.

If the distance between object and screen (D) is greater than 4f, then there are two positions of the lens for which the image of object on the screen is distinct and clear. Using sign convention for real image formed by a lens, we have

$\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u} \cdots Equation \, \, 1$

Clearly u and v are interchangeable, i.e. in these two positions the distances of object and image from the lens are interchanged.

I.e.

$v_1 = u_2 \, \, and v_2 = u_1$

The figure above represents the formation of real images of an object in two positions $P_1 \, \, and \, \, P_2$ of lens L when D > 4f.

If $I_1 \, \, and \, \, I_2$ are the sizes of images in I and II positions of lens L, O is the size of object and $m_1 , m_2$ magnifications produced by lens in I position, and II position respectively, then,

$m_1 = \dfrac{v_1}{u_1} = \dfrac{I_1}{O}$ and

$m_2 = \dfrac{v_2}{u_2} = \dfrac{I_2}{O}$

Multiplying (4) and (5), we get

$m_1 m_2 = \dfrac{v_1}{u_1} \times \dfrac{v_2}{u_2} = \dfrac{I_1 I_2}{o^2}$

But $u_2 = v_1 = v \, \, and \, \, v_2 = u_1 =u$

$\therefore m_1 m_2 = \dfrac{v}{u} \times \dfrac{u}{v} = \dfrac{I_1 I_2}{O^2}$

$\therefore Size \, \, of \, \, object \, \, O = \sqrt{I_1 I_2}$ and focal length of lens.

$\therefore f = \dfrac{D^2 - x^2}{4D} \cdots Equation \, \, 8$

Displacement method

Silvering of one surface of lens

If one of the surfaces of a lens is silvered, the rays are first refracted by lens, then reflected from silvered surface and finally refracted by lens, so that the effective focal length of lens is:

$\dfrac{1}{F} = \dfrac{1}{f_i} + \dfrac{1}{f_m} + \dfrac{1}{f_i} = \dfrac{2}{f_l} + \dfrac{1}{f_m}$

Where $f_l$ is focal length of lens and $f_m$ is focal length of spherical mirror of radius of curvature of silvered surface.

This lens acts as a concave mirror and so the formula for image formation will be $\dfrac{1}{F} = \dfrac{1}{v} + \dfrac{1}{u}$

Two thin lenses separated by a distance: If two thin lenses of focal lengths $f_1 , f_2$ are placed at a distance d apart, then equivalent focal length of combination is:

$\dfrac{1}{F} = \dfrac{1}{f_1} + \dfrac{1}{f_2} - \dfrac{d}{f_1 f_2}$

Or, power of combination $P = P_1 + P_2 - dp_1 p_2$

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