4 Engineering Example 4

4.1 Optical interference fringes due to a glass plate

Monochromatic light of intensity I 0 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 :

{ Geometry of a light ray transmitted and reflected through a glass plate}

The intensity I of the light wave transmitted through the plate is given by

I = I 0 t 4 1 + | r | 4 − 2 | r | 2 cos θ

where t and r are the complex transmission and reflection coefficients. The phase angle θ is the sum of

(i) a phase proportional to the incidence angle α 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 θ 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

I = I 0 t 4 1 + | r | 4 − 2 | r | 2 cos θ (1)

The light intensity depends solely on the variable θ as shown in equation (1), and the objective is to find the values θ that will minimize and maximize I . The angle θ is introduced in equation (1) through the function cos θ in the denominator. We consider first the maxima of I .

Light intensity maxima

I is maximum when the denominator is minimum. This condition is obtained when the factor 2 r cos θ is maximum due to the minus sign in the denominator. As stated in Section 4.2, the maxima of 2 r cos θ occur when cos θ = + 1. Values of cos θ = + 1 correspond to θ = 2 n π where n = … − 2 , − 1 , 0 , 1 , 2 , … (see Section 4.5) and θ is measured in radians. Setting cos θ = + 1 in equation (1) gives the intensity maxima

I max = I 0 t 4 1 + | r | 4 − 2 | r | 2 .

Since the denominator can be identified as the square of ( 1 + r 2 ) , the final result for maximum intensity can be written as

I max = I 0 t 4 ( 1 − r 2 ) 2 . (2)

Light intensity minima

I 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 2 r cos θ is minimum. The minima of 2 r cos θ occur when cos θ = − 1. Values of cos θ = − 1 correspond to θ = π ( 2 n + 1 ) where n = … − 2 , − 1 , 0 , 1 , 2 , … (see Section 4.5). Setting cos θ = − 1 in equation (1) gives an expression for the intensity minima

I min = I 0 t 4 1 + | r | 4 + 2 | r | 2 .

Since the denominator can be recognised as the square of ( 1 + r 2 ) , the final result for minimum intensity can be written as

I min = I 0 t 4 ( 1 + r 2 ) 2 (3)

Interpretation

The interference fringes for intensity maxima or minima occur at constant angle θ 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

  1. a fringe of maximum light at the centre (bright dot for θ = 0 ),
  2. a circular fringe of minimum light at angle θ = π ,
  3. a circular fringe of maximum light at 2 π etc.

Figure 32 :

{ Sketch of interference fringes due to a glass plate}

Exercises
  1. Express the following angles in radians (as multiples of π )
    1. 12 0 ∘
    2. 2 0 ∘
    3. 13 5 ∘
    4. 30 0 ∘
    5. − 9 0 ∘  
    6. 72 0 ∘
  2. Express in degrees the following quantities which are in radians
    1. π 2
    2. 3 π 2
    3. 5 π 6
    4. 11 π 9
    5. − π 8
    6. 1 π
  3. Obtain the precise values of all 6 trigonometric functions of the angle θ for the situation shown in the figure:

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  4. Obtain all the values of x between 0 and 2 π such that
    1. sin x = 1 2   
    2. cos x = 1 2
    3. sin x = − 3 2
    4. cos x = − 1 2
    5. tan x = 2
    6. tan x = − 1 2
    7. cos ( 2 x + 6 0 ∘ ) = 2
    8. cos ( 2 x + 6 0 ∘ ) = 1 2
  5. Obtain all the values of θ in the given domain satisfying the following quadratic equations
    1. 2 sin 2 θ − sin θ = 0 0 ≤ θ ≤ 36 0 ∘
    2. 2 cos 2 θ + 7 cos θ + 3 = 0 0 ≤ θ ≤ 36 0 ∘
    3. 4 sin 2 θ − 1 = 0
    1. Show that the area A of a sector formed by a central angle θ radians in a circle of radius r is given by

      A = 1 2 r 2 θ .

      (Hint: By proportionality the ratio of the area of the sector to the total area of the circle equals the ratio of θ to the total angle at the centre of the circle.)

    2. What is the value of the shaded area shown in the figure if θ is measured (i) in radians, (ii) in degrees?

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  7. Sketch, over 0 < θ < 2 π , the graph of
    1. sin 2 θ
    2. sin 1 2 θ
    3. cos 2 θ  
    4. cos 1 2 θ .

      Mark the horizontal axis in radians in each cas

    5. Write down the period of sin 2 θ and the period of cos 1 2 θ .
    1. 2 π 3
    2. π 9
    3. 3 π 4
    4. 5 π 3
    5. − π 2
    6. 4 π
    1. 1 5 ∘
    2. 27 0 ∘
    3. 15 0 ∘
    4. 22 0 ∘
    5. − 22 . 5 ∘
    6. 18 0 ∘ π 2
  3. The distance of the point P from the origin is r = ( − 3 ) 2 + 1 2 = 10 . Then, since P lies on a circle radius 10 rather than a circle of unit radius:
    sin θ = 1 10 cosec θ = 10 cos θ = − 3 10 sec θ = − 10 3 tan θ = 1 − 3 = − 1 3 cot θ = − 3
    1. x = 4 5 ∘ π 4 radians x = 13 5 ∘ 3 π 4 (recall sin ( 180 − x ) = sin x )
    2. x = 6 0 ∘ π 3 x = 30 0 ∘ 5 π 3
    3. x = 24 0 ∘ 4 π 3 x = 30 0 ∘ 5 π 3
    4. x = 13 5 ∘ 3 π 4 x = 22 5 ∘ 5 π 4
    5. x = 63.4 3 ∘ x = 243.4 3 ∘ (remember tan x has period 18 0 ∘ or π radians)
    6. x = 153.4 3 ∘ x = 333.4 3 ∘
    7. No solution !
    8. x = 0 ∘ , 12 0 ∘ , 18 0 ∘ , 30 0 ∘ , 36 0 ∘
    1. 2 sin 2 θ − sin θ = 0   so   sin θ ( 2 sin θ − 1 ) = 0   so   sin θ = 0

      giving θ = 0 ∘ , 18 0 ∘ , 36 0 ∘   or   sin θ = 1 2   giving   θ = 3 0 ∘ , 15 0 ∘

    2. 2 cos 2 θ + 7 cos θ + 3 = 0 . With x = cos θ we have 2 x 2 + 7 x + 3 = 0 ( 2 x + 1 ) ( x + 3 ) = 0  (factorising) so   2 x = − 1 or x = − 1 2 . The solution x = − 3 is impossible since x = cos θ .

      The equation x = cos θ = − 1 2 has solutions θ = 12 0 ∘ , 24 0 ∘

    3. 4 sin 2 θ = 1 so sin 2 θ = 1 4 i.e. sin θ = ± 1 2 giving θ = 3 0 ∘ , 15 0 ∘ , 21 0 ∘ , 33 0 ∘
    1. Using the hint,

      θ 2 π = A π r 2

      from where we obtain A = π r 2 θ 2 π = r 2 θ 2

    2. With θ in radians the shaded area is

      S = R 2 θ 2 − r 2 θ 2 = θ 2 ( R 2 − r 2 )

      If θ is in degrees, then since x radians = 180 x ∘ π or x ∘ = π x 180 radians, we have

      S = π θ ∘ 36 0 ∘ ( R 2 − r 2 )

  7. The graphs of sin 2 θ and cos 2 θ are identical in form with those of sin θ and cos θ respectively but oscillate twice as rapidly.

    The graphs of sin 1 2 θ and cos 1 2 θ oscillate half as rapidly as those of sin θ and cos θ .

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    From the graphs sin 2 θ has period 2 π and cos 1 2 θ has period 4 π . In general sin n θ has period 2 π ∕ n .