Optics Interference: Important Solved Problems

 

ANNA UNIVERSITY SOLVED PROBLEMS

 

1. An air‒wedge is formed using two glass plates each of length 5cm and a thin wire. The wavelength of light used is 589nm. If 200 fringes are formed, find the radius of the wire.

Given data

 λ = 589 × 10^‒9m; x = 5cm = 5 × 10^‒2m

Solution

 β = 5×10^‒2 / 200

Thickness or diameter of the wire

t = xλ / 2β

 = [5×10^‒2×589×10⁻⁹×200] / [2×5×10^‒2]

t = 589 × 10^‒7m = 58.9 μm

Diameter of the wire = 58.9 μm

Radius of the wire = 58.9 / 2 = 29.45 μm

Radius of the wire = 29.45 μm

 

2. Interference fringes are produced with monochromatic light falling normally on a wedge shaped film whose refractive index is 1.40. The angle of wedge is 20 seconds of an arc and the distance between the successive fringes is 0.25 cm. Calculate the wavelength of light used.

Given data

 θ = 20 seconds = [ 20π / 60×60×180 ] radians

 μ = 1.40

 β = 0.25 × 10^‒2m

Solution

Formula: Fringe width β = λ / 2μθ

 λ = 2μθβ

 λ = [2×1.4×20π × 0.25×10^‒2] / [60×60×180]

 λ = 0.4396 / 648000

= 6.783 × 10^‒7m

∴. Wavelength of light λ = 6783 Å

 

3. In the Michelson interferometer arrangement, if one of the mirrors is moved by 0.04mm, 125 fringes cross the field of view. Calculate the wavelength of light used.

Given data

d = 0.04mm = 0.04 × 10^‒3m

n = 125

Solution

Formula: Wavelength λ = 2d / n

λ = 2d / n

λ = 2×0.04×10^‒3 / 125

= 6.4 ×10^‒7 m

= 6400Å (Red in colour)

Wavelength of light used = 6400Å

 

4. In a Michelson's interferometer experiment 200 circular fringes cross the field of view when the movable mirror is displaced through a distance 0.0589 mm. Calculate the wavelength of the monochromatic source used.

Given data

n = 200 fringes,

d = 0.0589×10^‒3 m

Solution

Wavelength λ = 2d / n

λ = 2×0.0589×10^‒3 / 200

λ = 5.89×10^‒7 m

λ = 5890 Å (yellow)

Wavelength of monochromatic source = 5890 Å

 

5. An air wedge is formed between two glass plates of length 10 cm each, with a fine insulated wire kept between the plates at one end. When illuminated with sodium light of wavelength 589.3 nm, 100 fringes are seen in 1 cm. With the insulation removed, 60 fringes are seen in 1 cm. Calculate the diameter of the wire and the thickness of insulation coating.

Solution

l = 10 cm;

λ = 5893 Å

Given

With insulation:

Fringe width β₁= 1×10^‒2 / 100 = 0.01×10^‒2m

Thickness with insulation is t₁ = lλ / 2β₁

 = 10×10²×5893×10^‒10 / 2×0.01×10^‒2

t₁ = 2.9465 μm

Without insulation:

Fringe width β₂= 1×10^‒2 / 60 = 0.01667×10^‒2m

Thickness without insulation,

i.e., the diameter of the wire alone is t₂ = lλ / 2β₂

= 10×10²×5893×10^‒10 / 2×0.01667×10^‒2

t₂ = 1.7675 μm

Diameter of the wire (t₂) =  1.7675 μm

∴ Thickness of the insulation (t₁ ‒ t₂) = 1.179 μm

Diameter of the wire = 1.7675 μm & Thickness of the insulation = 1.179 μm

 

6. When a thin film of glass of refractive index 1.5 is interposed in the path of one of the interfering beams of Michelson's interferometer, shift of 30 fringes of sodium light is observed across the field of view. If the thickness of the glass film is 0.018 mm, calculate the wavelength of the light used.

Solution

From the theory of interference, the path difference 2 (μ‒1)t =  nλ

2 (μ‒1)t =  nλ

  λ = 2 (μ‒1)t  /  n

(or) λ = 2(1.5−1)(0.018×10⁻³)  /  30

 λ = 6×10⁻⁷ m

Wavelength of the light (λ) = 6×10⁻⁷ m

 

 

ADDITIONAL SOLVED PROBLEMS

 

1. Monochromatic light of wavelength 5893Å incident normally on a wedge shaped film of refractive index 1.5, Interference fringes are formed that are 2mm apart. Find the angle of the wedge.

Solution

λ = 5893Å = 5893×10^‒10m

 μ = 1.5,

β = 2mm = 2 × 10^‒3m

β = λ / 2μθ

θ = λ / 2μβ

θ = 5893×10^‒10 / 2×1.5×2×10^‒3

θ = 0.982 × 10^‒4 radians.

The angle of the wedge= 0.982 × 10^‒4 radians.

 

2. Interference fringes are produced by monochromatic light of wavelength 5238Å incident normally on a wedge shaped film of glass of refractive index 1.5. The distance between the successive fringes is 1mm. Calculate the angle of the wedge in seconds of an arc.

Solution

 λ=5238Å = 5238 × 10^‒10m

μ=1.5

β=1mm = 1 × 10^‒3m

β = λ / 2μθ

θ = λ / 2μβ

θ = 5893×10^‒10 / 2×1.5×1×10^‒3 radians

θ = 5893×10^‒10×180 / 2×1.5×1×10^‒3×π

θ = 5893×10^‒10×180×60×60  /  2×1.5×1×10^‒3×3.14 seconds of an arc.

The angle of the wedge θ = 36.03 seconds of an arc.

 

3. A beam of monochromatic light of wavelength 5800Å falls on the wedge shaped air film with an angle 0.02 radian. At what distance from the edge of the wedge, will the 8th fringe be observed.

Solution

 λ = 5800 × 10^‒10 m

 θ = 0.02 radian

n= 8 and

μ = 1

β = nλ / 2μθ

Also we know

 β = x / n

λ / 2μθ = x / n

x = nλ / 2μθ

=  8×5800×10^‒10 / 2×0.02×1

= 1.16×10^‒4 m

The distance from the edge of the wedge = 1.16×10^‒4 m

 

4. Monochromatic light of wavelength 5840Å from a source falls normally on the two glass plates enclosing a wedge shaped airfilm. The plates are separated by a wire of 0.03mm diameter at a distance of 18cm from the edge. Find the fringe width.

Solution

 x = 18 cm

 λ = 5840 × 10^‒10 cm

 μ = 1

Diameter (or) Thickness (t) = 0.03 mm or 0.003 cm

 θ = t/x  = 0.03×10^‒3 / 18×10^‒2

= 1.666×10^‒4 radians.

β = λ / 2μθ

β = 5893×10^‒10 / 2×1×1.66×10^‒4

 = 1.752×10^‒3 m

Fringe width β = 1.752×10^‒3 m

 

5. A beam of monochromatic light of wavelength 5820Å is incident on a glass wedge. The wedge angle is 20 seconds of an arc and the refractive index is the made glass is 1.5. Find the number of dark fringes per cm of the wedge length.

Solution

 θ = 20 seconds of an arcon

λ = 5820 × 10^‒10 m

μ = 1.5

β = λ / 20μ

β = 5820×10^‒10×60×60×180 / 2×20×π×1.5

β = 0.2 cm

∴ Number of fringes per cm (x) = x/β = 1/0.2 = 5 per cm

 

6. A wedge shaped film of glass of refractive index 1.42 and an angle of wedge is 25 seconds of an arc is illuminated by the monochromatic source. If the distance between the successive fringes is 0.2 cm, calculate the wavelength of the light.

Solution

 θ = 25 seconds of an arc

 = 25×π / 60×60×180 radians

 β = 0.3 cm

 μ = 1.42

β = λ / 2θμ

λ = 2θμβ

λ = 2×25×π×1.42×0.2 / 60×60×180

λ = 6880 × 10^‒8 cm

∴ The wavelength of the light λ = 6880Å

 

7. When the wedge shaped air film between the two plates is illuminated by the monochromatic source of wavelength 5800Å, 15 fringes per cm is formed. Calculate the wedge angle in seconds.

Solution

 x=1,

n=15 fringes,

λ =5800 × 10^‒10 m

 β = x/n = 1/15 cm (or)

= 1/1500 m

 β=λ/2θ

θ = λ / 2β

= 5800×10^‒10×1500 / 2×1

= 4.35 × 10^‒4 radian

θ = 4.35×10^‒4×180×60×60 / π    seconds of an arc

The wedge angle = 89.7 seconds of an arc.

 

8. A shift of 450 fringes is observed when the movable mirror of Michelson interferometer is shifted by 0.1572 mm. Calculate the wavelength of the light.

Solution

n = 450

d= 0.1572 mm

Formula: d = nλ / 2

λ = 2d / n

= 2×0.1572×10^‒3 / 450

λ = 6986×10^‒10 m

The wavelength of the light = 6986Å

 

9. Fringes of equal inclination are observed in a Michelson interferometer. As one minor is moved by 1 mm, 3650 fringes are cross the field of view. Calculate wavelength of light.

Solution

n = 3650,

d=1mm (or) 1× 10^‒3 m

Formula: 2d=nλ

λ = 2d / n

= 2×1×10^‒3 / 3650

= 5479Å

The wavelength of the light = 5479Å

 

10. In a Michelson interferometer, when a thin film of refractive index 1.5 is interposed in the path of one of the interfering beams, a shift of 30 fringes is observed in the field of view. If the thickness of the thin film is 18 μm, calculate the wavelength of the light source used in the experiment.

Solution

Given data

μ=1.5,

n = 30,

t = 18× 10^‒6 m

Formula: 2(μ‒1) t = nλ

2(μ‒1) t = nλ

 λ = 2(μ‒1) t  /  n

= 2(1.5‒1)×18×10^‒6 / 30

= 6×10^‒7m (or) 6000 Å

Wavelength of the light source = 6000 A

 

11. When thin film of glass is introduced in one of the arms of the Michelson interferometer, it displaces 20 fringes. If the wavelength of the light used is 5890Å and the refractive index of the thin plate is 1.48. Calculate the thickness of the plate.

Solution

λ = 5890 × 10^‒10m

μ = 1.48,

n = 20

Formula: 2(μ‒1) t = nλ

 t = nλ / 2(μ‒1)

= 20×5890×10^‒10 / 2(1.48 − 1)

= 1.227 × 10^‒5 m

The thickness of the plate = 1.227 × 10^‒5 m

 

 

ASSIGNMENT PROBLEMS

 

1. Monochromatic light of wavelength 5900Å is incident normally on a wedge shaped film of cellophane whose refractive index is 1.42. If the interference fringes are obtained with the fringe width 0.26 mm, find the angle of the wedge. (Ans: 7.99 × 10^‒4 radian)

 

2. Interference fringes are obtained by monochromatic source of light falling normally on the thin wedge shaped film of refractive index 1.45. If the angle of the wedge is 18 seconds of an arc and the distance between the successive fringes is 0.2cm. Calculate the wavelength of light. (Ans: 5063Å)

 

3. If the angle of wedge is 0.3° and the wavelength of the Sodium lines are 5890Å & 5896Å. Find the distance from the apex at which the maxima due to two wavelengths first coincide when observed in reflected light. (Ans: 5.541 cm)

 

4. A square piece of cellophane film with refractive index 1.45 has wedge shaped section, so that thickness at the two opposite sites are t₁ and t₂. If the wavelength of the light is 5890Å is the number of fringes appearing on the foil is 10 calculate the difference (t₂ ‒ t₁)  (Ans: 2.031 × 10^‒6 m)

 

5. On introducing a thin film of glass of refractive index 1.5 and thickness 0.00345 cm in the path of one of the interfering beam of Michelson interferometer a certain number of fringes passes across the field of view. If the wavelength of the light used is 6900Å. Calculate the number of fringes passed across the field of view. (Ans: 50)