4 Engineering Example 4
4.1 Optical interference fringes due to a glass plate
Monochromatic light of intensity propagates in air before impinging on a glass plate (see Figure 31). If a screen is placed beyond the plate then a pattern is observed including alternate light and dark regions. These are interference fringes.
Figure 31 :
The intensity of the light wave transmitted through the plate is given by
where and are the complex transmission and reflection coefficients. The phase angle is the sum of
(i) a phase proportional to the incidence angle and
(ii) a fixed phase lag due to multiple reflections.
The problem is to establish the form of the intensity pattern (i.e. the minima and maxima characteristics of interference fringes due to the plate), and deduce the shape and position of the fringes captured by a screen beyond the plate.
Solution
The intensity of the optical wave outgoing from the glass plate is given by
(1)
The light intensity depends solely on the variable as shown in equation (1), and the objective is to find the values that will minimize and maximize . The angle is introduced in equation (1) through the function in the denominator. We consider first the maxima of
Light intensity maxima
is maximum when the denominator is minimum. This condition is obtained when the factor is maximum due to the minus sign in the denominator. As stated in Section 4.2, the maxima of occur when Values of correspond to where (see Section 4.5) and is measured in radians. Setting in equation (1) gives the intensity maxima
Since the denominator can be identified as the square of , the final result for maximum intensity can be written as
(2)
Light intensity minima
is minimum when the denominator in (1) is maximum. As a result of the minus sign in the denominator, this condition is obtained when the factor is minimum. The minima of occur when Values of correspond to where (see Section 4.5). Setting in equation (1) gives an expression for the intensity minima
Since the denominator can be recognised as the square of the final result for minimum intensity can be written as
(3)
Interpretation
The interference fringes for intensity maxima or minima occur at constant angle and therefore describe concentric rings of alternating light and shadow as sketched in the figure below. From the centre to the periphery of the concentric ring system, the fringes occur in the following order
- a fringe of maximum light at the centre (bright dot for ),
- a circular fringe of minimum light at angle ,
- a circular fringe of maximum light at etc.
Figure 32 :
Exercises
-
Express the following angles in radians (as multiples of
)
-
Express in degrees the following quantities which are in radians
-
Obtain the
precise
values of all 6 trigonometric functions of the angle
for the situation shown in the figure:
-
Obtain all the values of
between 0 and
such that
-
Obtain all the values of
in the given domain satisfying the following quadratic equations
-
-
Show that the area
of a sector formed by a central angle
radians in a circle of radius
is given by
(Hint: By proportionality the ratio of the area of the sector to the total area of the circle equals the ratio of to the total angle at the centre of the circle.)
-
What is the value of the shaded area shown in the figure if
is measured (i) in radians, (ii) in degrees?
-
Show that the area
of a sector formed by a central angle
radians in a circle of radius
is given by
-
Sketch, over
, the graph of
-
.
Mark the horizontal axis in radians in each cas
- Write down the period of and the period of .
-
-
-
The distance of the point
from the origin is
. Then, since
lies on a circle radius
rather than a circle of unit radius:
-
- (recall )
- (remember has period or radians)
- No solution !
-
-
so
so
giving or giving
-
. With
we have
(factorising) so
or
. The solution
is impossible since
.
The equation has solutions
- so i.e. giving
-
so
so
-
-
Using the hint,
from where we obtain
-
With
in radians the shaded area is
If is in degrees, then since radians or radians, we have
-
Using the hint,
-
The graphs of
and
are identical in form with those of
and
respectively but oscillate twice as rapidly.
The graphs of and oscillate half as rapidly as those of and .
From the graphs has period and has period . In general has period .