Subject: Physics
When a wave passes through a medium, the particles of the medium vibrate. When the particles completer one to and fro motion, the wave advances by a distance equal to its wavelength λ. For a complete wave, the wavelength varies in λ and the phase is changed through 2p.
Let there be two waves with a path difference of λ. Then, the phase difference them will be 2p. If the path difference is x, then path difference \( = \frac {2\pi }{\lambda } \times x. \)
$$ \text {hence, phase difference} = \frac {2\pi }{\lambda } \times \text {path difference} $$
The product of the distance travelled by the light in a medium and the refractive index of that medium is called the optical path. If d be the distance travelled by the light in a medium of refractive index µ, then by definition,
$$ \text {Optical Path} = \mu d $$
Let us consider an optically denser medium in which the light travels with a velocity v. if the distance travelled by the light in the medium is d, then the time taken to ravel this distance is given by
$$ t = \frac dv \dots (i) $$
In this time, the light travels a distance L in free space. Then,
$$ L = ct \dots (ii) $$
Now, the refractive index of the medium is given by
$$ \mu = \frac cv \dots (iii) $$
where c is the velocity of the light in a free space. Substituting the value of t from equation (i) in equation (ii), we get
\begin{align*} L = c \times \frac dv = \frac {cv}{v} \\ \text {or,} \: L = \mu d \\ \end{align*}
This is called the optical path.
Conditions for Sustained Interference of Light
S is a narrow vertical slit (of width about 1 mm) illuminated by a monochromatic source of light. At a suitable distance (about 10 cm ) from S, there are two fine slits S1 and S2 about 0.5 mm apart at equidistant from S. when a screen is placed at a larger (about 2m) from the slits S1 and S2, alternate bright and dark bands appear on the screen. The appearance of bright and dark bands are called the fringes.
Suppose S1 and S2 be two fine slits at a small distance d apart in the figure. Let slits are illuminated by monochromatic light from a strong source S of wavelength λ and MN is a screen at a distance D from the double slits. The two waves starting from S1 and S2 superimpose upon each other resulting an interference pattern on the screen placed parallel to the double slit as in the figure.
Let O be the centre between the slits S1 and S2. Draw S1P, S2P and OC perpendicular to MN. The intensity of light at a point on the screen will depend upon the path difference between the two waves arriving at the point. The point C on the screen lies on the perpendicular bisector of S1 and S2. Therefore, the path difference between two waves reaching C is zero and hence, they are in phase. So, the point C is the position of maximum intensity. It is called central maximum.
Consider a point P at a distance x from C. The path difference between two waves arriving at P is given by
$$\text {path difference} = S_2P – S_1P $$
From the geometry in figure, it is found that
\begin{align*} PQ = x-\frac d2 ; PR = x + \frac d2 \\ \text {And} (S_2P)^2 –(S_1P)^2 = \left [ D^2 + \left ( x + \frac d2 \right )^2 \right ] - \left [ D^2 + \left ( x - \frac d2 \right )^2 \right ]\\ \text {or,} \: (S_2P – S_P)(S_2P + S_1P) = 2xd \\ \text {or,} \: (S_2P – S_P)= \frac {2xd}{BP + AP} \\ \text {In practice, point P lies very close to C.} \\ \text {So} \: S_2P \approx S_1P \approx D \\ S_2P + S_1P = D + D = 2D \\ \text {Path difference} = S_2P – S_1P = \frac {2xd}{2D} = \frac {xd}{D} \\ \end{align*}
The waves from S1 and S2 arriving at a point on the screen will interfere constructively or destructively depending upon this path difference. The phase difference for this path difference is given by
$$ \text {Phase difference,} \: \phi = \frac {2\pi}{\lambda } \left (\frac {xd}{D} \right )$$
Reference
Manu Kumar Khatry, Manoj Kumar Thapa,et al. Principle of Physics. Kathmandu: Ayam publication PVT LTD, 2010.
S.K. Gautam, J.M. Pradhan. A text Book of Physics. Kathmandu: Surya Publication, 2003.
When the particles completer one to and fro motion, the wave advances by a distance equal to its wavelength λ.
The product of the distance travelled by the light in a medium and the refractive index of that medium is called the optical path.
The appearance of bright and dark bands are called the fringes.
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