In this article, we shall study numerical problems based on Young’s experiment and biprism experiment to find the fringe width of the interference pattern and to find the wavelength of light used.

Example – 12:

In a biprism experiment, the slit is illuminated by the light of wavelength 5890 Å. The distance between the slit and the eye-piece is 0.8 m. The two virtual images of the slit are formed 0.2 cm apart. Calculate the change in fringe width if the eye-piece is displaced 10 cm away from the slit.

Solution:

Part – I:

Given: Distance between slits = d = 0.2 cm = 0.2 x 10^-2 m = 2 x 10^-3 m. Distance between slit and screen = D = 0.8 m, Wavelength of light = λ = 5890 Å = 5890 x 10^-10 m = 5.89 x 10^-7 m

To Find: Fringe width = X =?

The fringe width is given by X = λD/d

X = λD/d = (5.89 x 10^-7 x 0.8)/( 2 x 10^-3) = 2.356 x 10^-4 m = 0.2356 mm

Part – II:

Given: For first case: fringe width = X₁ = 0.2356 mm, distance of eye piece from the slits = D₁ = 0.8 m, For second case: distance of eye piece from the slits = D₂ = 0.8 m  + 10 cm = 0.8 m + 0.1 m = 0.9 m

To Find: Change fringe width = ∆X =?

For first case X₁ = λD₁/d ………….. (1)

For second case, X₂ = λD₂/d ………….. (2)

Dividing equation (2) by (1)

X₂ / X₁ = (λD₂/d) x (d/ λD₁)

∴ X₂ / X₁ = D₂/D₁

∴ X₂ = (D₂/D₁) x X₁

∴ X₂ = (0.9/0.8) x 0.2356 = 0.2651 mm

∴ Change in fringe width = ∆X = X₂ – X₁ = 0.2651 – 0.2356 = 0.0295 mm

Ans: The change in fringe width is 0.0295 mm

Example – 13:

In Young’s experiment the distance between two consecutive bright bands produced on a screen placed at 1.5 m from the two slits is 6.5 mm. What would be the fringe width if the screen is brought towards the slit by 50 cm for the same setting?

Given: For first case: fringe width = X₁ = 6.5 mm, distance of screen from the slits = D₁ = 1.5 m, For second case: distance of screen from the slits = D₂ = 1.5 m  – 50 cm = 1.5 m + 0.5 m = 1 m

To Find: New fringe width = X =?

Solution:

The fringe width is given by X = λD/d

For the first case X₁ = λD₁/d ………….. (1)

For the second case, X₂ = λD₂/d ………….. (2)

Dividing equation (2) by (1)

X₂ / X₁ = (λD₂/d) x (d/ λD₁)

X₂ / X₁ = D₂/D₁

X₂ = (D₂/D₁) x X₁

X₂ = (1/1.5) x 6.5 = 4.33 mm

Ans: New fringe width is 4.33 mm

Example – 14:

In a biprism experiment, the fringe width is 0.4 mm when the eye-piece is at a distance of one metre from the slit. If now only the eye-piece is moved 25 cm towards the biprism, what would be the change in fringe width?

Given: For first case: fringe width = X₁ = 0.4 mm, distance of eye piece from the slits = D₁ = 1 m, for second case: distance of eye piece from the slits = D₂ = 1 m – 25 cm = 1 m + 0.25 m = 0.75 m

To Find: Change fringe width = ∆X =?

Solution:

The fringe width is given by X = λD/d

For first case  X₁ = λD₁/d ………….. (1)

For second case, X₂ = λD₂/d ………….. (2)

Dividing equation (2) by (1)

X₂ / X₁ = (λD₂/d) x (d/ λD₁)

X₂ / X₁ = D₂/D₁

X₂ = (D₂/D₁) x X₁

X₂ = (0.75/1) x 0.4 = 0.3 mm

Change in fringe width = ∆X = X₂ – X₁ = 0.4 – 0.3 = 0.1 mm

Ans: There is decrease of 0.1 mm in fringe width

Example – 15:

In biprism experiment, light of wavelength 5200 Å is used to get an interference pattern on a screen. The fringe width changes by 1.3 mm when the screen is brought towards the biprism by 50 cm. Find the distance between the virtual images of the slit.

Given: Wavelength of light used = λ = 5200 Å = 5200 x 10^-10 m = 5.2 x 10^-7 m, change in fringe width = ΔX = 1.3 mm = 1.3 x 10^-3 m, Initial distance of screen from virtual image = D₁ = D m, final distance of screen from virtual image = D₂ = D m – 50 cm = (D – 0.5) m

To Find: distance between the virtual images of the slit = d =?

Solution:

The fringe width is given by X = λD/d

For first case X₁ = λD₁/d ………….. (1)

For second case, X₂ = λD₂/d ………….. (2)

∆X = X₁ – X₂ = λD₁/d – λD₂/d = (λ/d)(D₁ – D₂) = (λ/d)(D – (D – 0.5)) =(λ/d)(0.5)

The change in fringe width = ∆X = (λ/2d)

d = (λ/2∆X) = (5.2 x 10^-7) / (2 x1.3 x 10^-3) = 2 x 10^-4 m = 0.2 mm

Ans: The distance between the virtual images of the slit is 0.2 mm

Example- 16:

In biprism experiment, the separation of the slits is halved and the distance between the slits and the screen is doubled. How is the fringe width affected?

Given: Distance between slits d₂ = ½ d₁, Distance between screen and slit D₂ = 2 D₁

To Find: Change in fringe width = ∆X =?

Solution:

The fringe width is given by X = λD/d

For first case X₁ = λD₁/d₁ ………….. (1)

For second case, X₂ = λD₂/d₂ ………….. (2)

Dividing equation (2) by (1)

X₂ / X₁ = (λD₂/d₂) x (d₁/ λD₁)

X₂ / X₁ = (D₂d₁)/( D₁d₂)

X₂ / X₁ = (D₂/D₁) x (d₁/d₂)

X₂ / X₁ = (2D₁/D₁) x (d₁/1/2 d₁) = 2 x 2 = 4

X₂ = 4 X₁

Ans: The fringe width increased 4 times

Example – 17:

In biprism experiment, interference fringes are observed at a distance of 60 cm from slits illuminated by monochromatic beam of light of wavelength 5460 Å. The distance between the slits is 3 mm. Find the change in fringe width if the distance between the slits is i) increased and ii) decreased by 1 mm.

Given: Distance between screen and sources = 60 cm = 0.6 m, Wavelength of light used = λ = 5460  Å = 5460 x 10^-10 m = 5.46 x 10^-7 m, Distance between slis = d = 3 mm = 3 x 10^-3 m.

Initial Fringe width = X = λD/d = (5.46 x 10^-7 x 0.6) / ( 3 x 10^-3) = 1.092 x 10^-4 m = 0.1092 mm

Part – I: The distance between slits is increased by 1 mm

Given: Initial fringe width = X₁ = 0.1092 mm, Initial distance between slits = d₁ = 3 mm, New distance between slits = d₂ = 3 + 1 = 4mm

To Find: the change in fringe width = ∆X =?

For first case X₁ = λD₁/d₁ ………….. (1)

For second case, X₂ = λD₂/d₂ ………….. (2)

Dividing equation (2) by (1)

X₂ / X₁ = (λD₂/d₂) x (d₁/ λD₁)

X₂ / X₁ = (D₂d₁)/( D₁d₂)

X₂ / X₁ = d₁/d₂

X₂ = (d₁/d₂) x X₁

X₂ = (3/4) x 0.1092 = 0.089

∆X = X₁ – X₂ =  0.1092 – 0.089 = 0.0273 mm

Part – Ii: The distance between slits is decreased by 1 mm

Given: Initial fringe width = X₁ = 0.1092 mm, Initial distance between slits = d₁ = 3 mm, New distance between slits = d₂ = 3 – 1 = 2mm

To Find: the change in fringe width = ∆X =?

For first case X₁ = λD₁/d₁ ………….. (1)

For second case, X₂ = λD₂/d₂ ………….. (2)

Dividing equation (2) by (1)

X₂ / X₁ = (λD₂/d₂) x (d₁/ λD₁)

X₂ / X₁ = (D₂d₁)/( D₁d₂)

X₂ / X₁ = d₁/d₂

X₂ = (d₁/d₂) x X₁

X₂ = (3/2) x 0.1092 = 0.1638

∆X = X₂ – X₁ =  0.1683 – 0.1092 = 0.0546 mm

Ans: When the distance between slits is increased by 1 mm, change in fringe width is 0.0273 mm. When the distance between slits is decreased by 1 mm, change in fringe width is 0.0546 mm

Example – 18:

In a biprism experiment, the distance of the slits from the screen is increased by 10% and the separation of slits is decreased by 20%. Find the percentage change in fringe width.

Given: Initial distance between the screen and slits = D₁, New distance of screen from slits = D₂ = D₁ + 10%D₁ = 1.10 D₁. Initial distance between the slits = d₁, New distance between the slits = d₂ = d₁ – 20%d₁ = 0.80 d₁

For first case X₁ = λD₁/d₁ ………….. (1)

For second case, X₂ = λD₂/d₂ ………….. (2)

Dividing equation (2) by (1)

X₂ / X₁ = (λD₂/d₂) x (d₁/ λD₁)

X₂ / X₁ = (D₂d₁)/( D₁d₂)

X₂ / X₁ = (D₂/D₁) x (d₁/d₂)

X₂ / X₁ = (1.10 D₁/D₁) x (d₁/0.80d₁)

X₂ = (1.10/0.80) x X₁

X₂ = 1.375 X₁

Change in fringe width = ∆X = 1.375X₁ – X₁ = 0.375 X₁

% change in fringe width = (∆X/X₁) x 100 = (0.375 X₁/X₁) x 100 = 37.5

Ans: There is increase of 37.5 % in the fringe width

Example – 19:

In a biprism experiment, interference bands are observed at a distance of one metre from the slit. A convex lens put between the slit and the eye-piece gibes two images of slit 0.7 cm apart, the lens being 30 cm from the slit. Calculate the width of 10 bands if a light of wavelength 5892 Å is used.

Given: Wavelength of light used = 5892 Å = = 5892 x 10^-10 m = 5.892 x 10^-7 m, Distance between the screen and slit = D = 1m

To Find: Width of 10 bands = 10X =?

ΔS₁‘S₂‘ P and S₁S₂P are similar

d/0.3 = 0.7/0.7

d = 0.3 cm = 0.3 x 10^-3 m = 3 x 10^-4 m

Fringe width = X = λD/d = (5.892 x 10^-7 x 1)/( 3 x 10^-4) = 1.964 x 10^-4 m = 0.1964 mm

Width of 10 bands = 10 x 0.1964 = 1.964 mm

Ans: The width 10 bands is 1.964 m

Example – 20:

In a biprism experiment, a certain fringe width is observed when the green light of wavelength 5350 Å is used. The distance between the slit and the screen is 1.28 m. What should be the distance between the slit and the screen if the red light of wavelength 6400 Å. is used to get the same fringe width without disturbing the distance between the coherent sources?

Given: Initial wavelength = λ₁ = 5350 Å, Initial distance between the slit and screen = D₁ = 1.28 m, New wavelength = λ₂ = 6400 Å,

To Find: New distance between the slit and screen = D₂ =?

The fringe width is given by X = λD/d

For first case X₁ = λ₁D₁/d ………….. (1)

For second case, X₂ = λ₂D₂/d ………….. (2)

Given X₁ = X₂

λ₁D₁/d= λ₂D₂/d

λ₁D₁ = λ₂D₂

D₂ = (λ₁/λ₂) x D₁

D₂ = (5350/6400) x 1.28 = 1.07 m

Ans: The distance between the slit and the screen for the red light is 1.07 m

Example – 21:

In biprism experiment when the slit is illuminated by light of wavelength 5890 Å, twenty fringes occupied 2.3 cm on the screen. When this light is substituted by another monochromatic light, thirty fringes occupied 2.8 cm on the screen. Find the wavelength of this light.

Given: Initial wavelength = λ₁ = 5890 Å, Initial fringe width = X₁ = 2.3/20 cm, New fringe width = X₂ = 2.8/30 cm,

To Find: New wavelength = λ₂ =?

For first case X₁ = λ₁D/d ………….. (1)

For second case, X₂ = λ₂D/d ………….. (2)

Dividing equation (2) by (1)

X₂ / X₁ = (λ₂D/d) x (d/ λ₁D)

X₂ / X₁ = λ₂/λ₁

λ₂ = (X₂/X₁) x λ₁ = ((2.8/30)/(2.3/20)) x 5890

λ₂ = (X₂/X₁) x λ₁ = (5.6/6.9) x 5890 = 4780 Å

Ans: The new wavelength used is 4780 Å

Example – 22:

In a biprism experiment, fringes were obtained with a monochromatic source of light. The eye-piece was kept at a distance of 1 m from the slit and the bandwidth was measured. When another monochromatic source was used without disturbing the arrangement, the same bandwidth was obtained when the eye-piece was at 80 cm from the slit. Calculate the ratio of wavelengths of light from the two sources.

Given:  Initial distance between the slit and screen = D₁ = 1 m, New distance between the slit and screen = D₂ = 80 cm = 0.8 m

To Find: Ratio of wavelengths = λ₁/λ₂ =?

For first case X₁ = λ₁D₁/d ………….. (1)

For second case, X₂ = λ₂D₂/d ………….. (2)

Given X₁ = X₂

λ₁D₁/d= λ₂D₂/d

λ₁D₁ = λ₂D₂

λ₁/λ₂ = D₂/D₁  = 0.8/1 = 8/10 = 4/5

Ans: The ratio of wavelengths used is 4:5

Previous Topic: More Problems on Fringe Width and Change of Fringe Width

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Science > Physics > Interference of Light > Numerical Problems on Fringe Width

The post Numerical Problems on Change in Fringe Width appeared first on The Fact Factor.

In this article, we shall study numerical problems based on Young’s experiment and biprism experiment to find the fringe width of the interference pattern and to find the wavelength of light used.

Example – 01:

In Young’s experiment, the distance between the two slits is 0.8 mm and the distance of the screen from the slits is 1.2m. If the fringe width is 0.75 mm, calculate the wavelength of light.

Given: Distance between slits = d = 0.8 mm = 0.8 x 10^-3 m = 8 x 10^-4 m. Distance between slit and screen = D = 1.2 m, Fringe width = X = 0.75 mm = 0.75 x 10^-3 m = 7.5 x 10^-4 m.

To Find: Wavelength of light used = λ =?

Solution:

The fringe width is given by X = λD/d

∴ λ = Xd/D = (7.5 x 10^-4 x 8 x 10^-4)/1.2 = 5 x 10^-7 m = 5000 x 10^-10 m = 5000 Å

Ans: Wavelength of light used is 5000 Å

Example – 02:

In Young’s experiment, the distance between the two images of the sources is 0.6 mm and the distance between the source and the screen is 1.5 m. Given that the overall separation between 20 fringes on the screen is 3 cm, calculate the wavelength of light used.

Given: Distance between images = d = 0.6 mm = 0.6 x 10^-3 m = 6 x 10^-4 m. Distance between source and screen = D = 1.5 m, Fringe width = X = (3/20) cm = 0.15 cm = 0.15 x 10^-2 m = 1.5 x 10^-3 m.

To Find: Wavelength of light used = λ =?

Solution:

The fringe width is given by X = λD/d

∴ λ = Xd/D = (1.5 x 10^-4 x 6 x 10^-3)/1.5 = 6 x 10^-7 m = 6000 x 10^-10 m = 6000 Å

Ans: Wavelength of light used is 6000 Å

Example – 03:

The distance between two consecutive bright bands in Young’s experiment is 0.32 mm when the red light of wavelength 6400 Å is used. By how much will this distance change if the light is substituted by the blue light of wavelength 4800 Å with the same setting?

Given: For red light: fringe width = X_r = 0.32 mm, wavelength = λ_r = 6400 Å, For blue light: wavelength = λ_b = 4800 Å

To Find: Change in fringe width = ∆X =?

Solution:

The fringe width is given by X = λD/d

For red light X_r = λ_rD/d ………….. (1)

For blue light, X_b = λ_bD/d ………….. (2)

Dividing Equation (2) by (1)

X_b/ X_r = (λ_bD/d) x (d/ λ_rD)

∴ X_b/ X_r = (λ_b/λ_r)

∴ X_b = (λ_b/λ_r) x X_r = (4800/6400) x 0.32 = 0.24 mm

∴ ∆X = X_b  – X_r  = 0. 32 – 0.24 = 0.08 mm

Ans: The distance between two consecutive bright bands will change by 0.08 mm

Example – 04:

Calculate the fringe width in the pattern produced in a biprism experiment given that the wavelength of light employed is 6000 Å, distance between sources is 1.2 mm and distance between the source and the screen is 100 cm. what will be the change in fringe width if the entire apparatus is immersed in water for which refractive index is 4/3.

Part – I:

Given: Distance between sorces = d = 1.2 mm = 1.2 x 10^-3 m, Distance between sources and screen = D = 100 cm = 1 m, Wavelength of light = λ = 6000 Å = 6000 x 10^-10 m = 6 x 10^-7 m

To Find: Fringe width = X =?

X = λD/d = (6 x 10^-7 x 1) / (1.2 x 10^-3) = 5 x 10^-4 m = 0.5 mm

Part – II:

Given: Fringe width in air = X₁ = 0.5 mm, wavelength = λ₁ = 6000 Å, for the second medium refractive index = μ = 4/3

To Find: Change in fringe width = ∆X =?

Solution:

μ = 4/3 =λ₁ / λ₂

The fringe width is given by X = λD/d

For first medium   X₁ = λ₁D/d ………….. (1)

For second medium, X₂ = λ₂D/d ………….. (2)

Dividing Equation (1) by (2)

X₁/ X₂ = (λ₁D/d) x (d/ λ₂D)

∴ X₁/ X₂ = (λ₁/λ₂) = 4/3

∴ X₂ = (3/4) X₁ = (3/4) x 0.5 = 0.375 mm

∴ ∆X = X₁  – X_2r  = 0. 5 – 0.375 = 0.125 mm

Ans: The initial fringe width is 0.5 mm and the fringe width changes by 0.125 mm

Example – 05:

In a Young’s experiment, a source of light having wavelength 6500 Å is replaced by a source of light of wavelength 5500 Å. Find the change in fringe width if the screen is at a distance of 1 m from the two sources which are 1 mm apart.

Part – I:

Given: Distance between sorces = d = 1 mm = 1 x 10^-3 m, Distance between sources and screen = D = 100 cm = 1 m, Initial wavelength of light = λ₁ = 6500 Å = 6500 x 10^-10 m = 6.5 x 10^-7 m, final wavelength of light = λ₂ = 5500 Å = 5500 x 10^-10 m = 5.5 x 10^-7 m,

To Find: Change in fringe width = ∆X =?

The fringe width is given by X = λD/d

For first medium   X₁ = λ₁D/d ………….. (1)

For second medium, X₂ = λ₂D/d ………….. (2)

∴ ∆X = X₁  – X₂ = λ₁D/d – λ₂D/d = (D/d)( λ₁ – λ₂) = (1/1 x 10^-3)(6.5 x 10^-7  – 5.5 x 10^-7)

∴ ∆X = X₁  – X₂ = λ₁D/d – λ₂D/d = (D/d)( λ₁ – λ₂) = 10³ x (6.5 – 6.5) x 10^-7  = 1 x 10^-4 m = 0.1 mm

Ans: The fringe width changes by 0.1 mm

Example – 06:

In a biprism experiment, the width of a fringe is 0.75 mm when the eye-piece is at a distance of one metre from the slits. The eye-piece is now moved away from the slits by 50 cm. Find the fringe width.

Given: For first case: fringe width = X₁ = 0.75 mm, distance of eye piece from the slits = D₁ = 1 m, For second case: distance of eye piece from the slits = D₂ = 1 m  + 50 cm = 1m + 0.5 m = 1.5 m

To Find: New fringe width =?

Solution:

The fringe width is given by X = λD/d

For first case X₁ = λD₁/d ………….. (1)

For second case, X₂ = λD₂/d ………….. (2)

Dividing equation (2) by (1)

X₂ / X₁ = (λD₂/d) x (d/ λD₁)

∴ X₂ / X₁ = D₂/D₁

∴ X₂  = (D₂/D₁) x X₁

∴ X₂  = (1.5/1) x 0.75 = 1.125 mm

Ans: New fringe width is 1.125 mm

Example – 07:

In Young’s experiment the distance between the slits is 1 mm and fringe width is 0.60 mm when light of certain wavelength is used. When the screen is moved through 0.25 m the fringe width increases by 0.15 mm. What is the wavelength of light used?

Given: For first case: fringe width = X₁ = 0.60 mm = 0.60 x 10^-3 m, distance between the slits = d = 1mm = 1 x 10^-3 m, Distance of screen from slit D₁ = D m. For second case: fringe width = X₂ = 0.6 + 0.15 = 0.75 mm, as the fringe width is increasing the screen is moved away from the slits, distance of the screen from the slits = D₂ = D m + 0.25 m = (D + 0.25) m

To Find: New fringe width =?

Solution:

The fringe width is given by X = λD/d

For first case X₁ = λD₁/d ………….. (1)

For second case, X₂ = λD₂/d ………….. (2)

Dividing equation (2) by (1)

X₂ / X₁ = (λD₂/d) x (d/ λD₁)

∴ X₂ / X₁ = D₂/D₁

∴ 0.75 / 0.60= (D + 0.25)/D

∴ 5/4= (D + 0.25)/D

∴ 5D = 4D + 1

∴ D = 1 m

Now, X₁ = λD₁/d

∴ λ =(X₁d) /D = (0.60 x 10-3 x 1 x 10-3) /1 = 6 x 10^-7 m

∴ λ = 6000 x 10^-10 m = 6000 Å

Ans: Wavelength of light used is 6000 Å

Example – 08:

In a Young’s experiment, the width of a fringe is 0.12 mm when the screen is at a distance of 100 cm from the slits. The screen is now moved towards the slit by 25 cm. Find the fringe width.

Given: For first case: fringe width = X₁ = 0.12 mm, distance of eye piece from the slits = D₁ = 100 cm = 1 m, For second case: distance of eye piece from the slits = D₂ = 1 m  – 25 cm = 1m – 0.25 m = 0.75 m

To Find: New fringe width =?

Solution:

The fringe width is given by X = λD/d

For first case X₁ = λD₁/d ………….. (1)

For second case, X₂ = λD₂/d ………….. (2)

Dividing equation (2) by (1)

X₂ / X₁ = (λD₂/d) x (d/ λD₁)

∴ X₂ / X₁ = D₂/D₁

∴ X₂ = (D₂/D₁) x X₁

∴ X₂ = (0.75/1) x 0.12 = 0.09 mm

Ans: New fringe width is 0.09 mm

Example – 09:

In Young’s experiment, interference bands are produced on the screen placed 1.5 m from two slits 0.15 mm apart and illuminated by the light of wavelength 6000 Å Find (1) fringe width (2) change in fringe width if the screen is taken away from the slits by 50 cm.

Solution:

Part – I:

Given: Distance between slits = d = 0.15 mm = 0.15 x 10^-3 m = 1.5 x 10^-4 m. Distance between slit and screen = D = 1.5 m, Wavelength of light = λ = 6000 Å = 6000 x 10^-10 m = 6 x 10^-7 m

To Find: Fringe width = X =?

X = λD/d = (6 x 10^-7 x 1.5) / ( 1.5 x 10^-4) = 6 x 10^-3 m = 6 mm

Part – II:

Given: For first case: fringe width = X₁ = 6 mm, distance of eye piece from the slits = D₁ = 1.5 m, for second case: distance of eye piece from the slits = D₁ = 1.5 m + 50 cm = 1.5 m + 0.5 m = 2 m

To Find: Change fringe width = ∆X =?

The fringe width is given by X = λD/d

For first case X₁ = λD₁/d ………….. (1)

For second case, X₂ = λD₂/d ………….. (2)

Dividing equation (2) by (1)

X₂ / X₁ = (λD₂/d) x (d/ λD₁)

∴ X₂ / X₁ = D₂/D₁

∴ X₂ = (D₂/D₁) x X₁

∴ X₂ = (2/1.5) x 6 = 8 mm

∴ Change in fringe width = ∆X = X₂ – X₁ = 8 – 6 = 2mm

Ans: Initial fringe width is 6 mm and the change in fringe width is 2 mm

Example – 10:

In Young’s experiment, interference bands are produced on the screen placed 1.5 m from two slits 1.5 mm apart and illuminated by the light of wavelength 4500 Å Find (1) fringe width (2) change in fringe width if the screen is moved towards the slits by 50 cm.

Solution:

Part – I:

Given: Distance between slits = d = 1.5 mm = 1.5 x 10^-3 m = 1.5 x 10^-4 m. Distance between slit and screen = D = 1.5 m, Wavelength of light = λ = 4500 Å = 4500 x 10^-10 m = 4.5 x 10^-7 m

To Find: Fringe width = X =?

X = λD/d = (4.5 x 10^-7 x 1.5) / (1.5 x 10^-3) = 4.5 x 10^-4 m = 0.45 mm

Part – II:

Given: For first case: fringe width = X₁ = 0.45 mm, distance of eye piece from the slits = D₁ = 1.5 m, For second case: distance of eye piece from the slits = D₁ = 1.5 m –  50 cm = 1.5 m – 0.5 m = 1 m

To Find: Change fringe width = ∆X =?

The fringe width is given by X = λD/d

For first case X₁ = λD₁/d ………….. (1)

For second case, X₂ = λD₂/d ………….. (2)

Dividing equation (2) by (1)

X₂ / X₁ = (λD₂/d) x (d/ λD₁)

∴ X₂ / X₁ = D₂/D₁

∴ X₂ = (D₂/D₁) x X₁

∴ X₂ = (1/1.5) x 0.45 = 0.30 mm

∴ Change in fringe width = ∆X = X₂ – X₁ = 0.45 – 0.30 = 0.15 mm

Ans: Initial fringe width is 0.45 mm and the change in fringe width is 0.15 mm

Example – 11:

In Young’s experiment, the fringe width is 0.65 mm when the screen is at a distance of 1.5 m from the slits. What will be the fringe width if the screen is moved towards the slits by 50 cm.

Solution:

Given: For first case: fringe width = X₁ = 0.65 mm, distance of eye piece from the slits = D₁ = 1.5 m, For second case: , distance of eye piece from the slits = D₁ = 1.5 m  –  50 cm = 1.5 m – 0.5 m = 1 m

To Find: fringe width = X₂ =?

The fringe width is given by X = λD/d

For first case X₁ = λD₁/d ………….. (1)

For second case, X₂ = λD₂/d ………….. (2)

Dividing equation (2) by (1)

X₂ / X₁ = (λD₂/d) x (d/ λD₁)

∴ X₂ / X₁ = D₂/D₁

∴ X₂ = (D₂/D₁) x X₁

∴ X₂ = (1/1.5) x 0.65 = 0.43 mm

Ans: New fringe width is 0.43 mm

Previous Topic: Concept of Fringe Width and Path Difference

Next Topic: More Problems on Fringe Wdth and Change of Fringe Width

Next Topic:

Science > Physics > Interference of Light > Numerical Problems on Fringe Width

The post Numerical Problems on Fringe Width appeared first on The Fact Factor.

In this article, we shall study Young’s double-slit experiment to demonstrate interference of light and also derive expressions for path difference and fringe width.

Young’s Double Slit Experiment:

The Young’s double-slit experimental arrangement consists of a pinhole ‘S’ punched on cardboard arranged in a dark room.  The pinhole is illuminated by a beam of sunlight.  The emergent light is allowed to fall on two more symmetrical pin holes S1 and S2 close to each other.  These two pinholes act as two coherent sources and give rise to secondary spherical wavefronts. The light emerging from S1 and S2 produces a steady interference pattern on a screen kept at a large distance away from them.  The interference pattern consists of alternately bright and dark ‘bands.  As sunlight consists of a large number of wavelengths the interference bonds are coloured as indistinct.

A modified Young’s experiment known as Young’s double-slit experiment gives rise to a sharp and clear Interference pattern.  In this, sunlight is replaced by a monochromatic source of light and pin holes are replaced by narrow slits parallel to each other. The interference pattern obtained on the screen consists of well-defined alternate bright and dark bands.  These bands are of equally spaced and are called interference bands or fringes.

Significance of Young’s Double Slit Experiment:

• Double slit experiment was the first experiment to demonstrate interference of light
• It gave the experimental confirmation of the wave nature of light.

Expression for Fringe Width of Interference Band:

Let S₁ and S₂ be two coherent thin sources (in form of slits) of monochromatic light. ‘d’ be the distance between them. Let ‘D’ be the distance of a screen from the sources kept parallel to the plane of the coherent sources; (D is large compared to d). Light waves emitted from the sources interfere and produce a stationary interference pattern on the screen which consists of alternately bright and dark fringes.  Let OC be the perpendicular bisector of S₁S₂ and P be a point on the screen such that CP = x. Draw S₁E and S₂F perpendicular to the screen.

For point C, x = 0   Thus, path difference = 0; so the point B will be a bright point. Hence interference fringe situated at C will be a bright fringe known as the central bright fringe. On either side of central bright fringe alternate dark and bright fringes will be situated. This set of bright and dark fringes is called an interference pattern.

Subtracting equation (2) from (1)

But S₂P – S1P is the path difference between the light waves reaching point P.
Hence path difference = xd/D …… (3)

Now we know that the brightness or darkness of a point depends upon the path difference. If path difference = n λ, where n = 0,1,2,3,….. the point will be bright and if path difference = (2n – 1)( λ /2), where n = 1,2,3,…. the point will be dark.

The distance between two consecutive dark bands or consecutive bright bands in an interference pattern is called the fringe width.

For Bright Fringe:

For bright band Path difference = n λ

∴  xd/ D = n λ

Let x_n be the distance of nth bright fringe from the central bright fringe. Let x_n+1  be the distance of (n + 1)th bright fringe from the central bright fringe.

But we know that the fringe width is the distance between two consecutive bright (or dark) fringes. Let X be the fringe width.

For Dark Fringe :

For dark band path difference = (2n – 1)( λ /2)

Let x_m be the distance of mth dark fringe from the central bright fringe. Let x_m+1 be the distance of (m + 1)th dark fringe from the central bright fringe.

From equation (5) and (6) we can conclude that the distance between two consecutive bright bands is the same as the distance between two consecutive dark bands. Thus the bright band and dark band are alternate and are equally spaced.

Fresnel’s Biprism experiment:

Principle :

Fresnel’s Biprism method gives rise to a well-spaced and sharp interference pattern. A biprism consists of a triangular prism having a very large refracting angle (or two identical wedge-shaped prisms of small acute angle joined with their bases in contact).

The biprism is held symmetrically with its refracting edge parallel to the slit.  The slit is illuminated with monochromatic light of wavelength λ. The divergent beam of light from the slit is incident on the biprism.  One biprism then produces two virtual images S1 and S2 of the slit S, which then acts as coherent sources of light at a small distance ‘d’ apart.  The interference fringes are obtained in the focal plane of the micrometer eyepiece E.

Apparatus:

The biprism experiment is performed by using an optical bench.  It consists of a heavy metal platform of about 2 metre long and provided with a scale along its length. The optical bench is fitted with four adjustable vertical stands which can be moved along the length of the bench. The stands support a slit, biprism, convex lens, and a micrometer eyepiece.

Adjustments:

The slit S is made narrow and is illuminated by light from a monochromatic source of wavelength λ kept behind it.  The light emerging from the slit is allowed to fall on the biprism B with its edge parallel to the slit.

The eyepiece. is kept at a sufficiently large distance D from the slit.  The slit, refracting edge of the biprism and the crosswire of the eyepiece are arranged parallel to each other in the same straight line.  By looking through the eyepiece, the biprism is slowly rotated about a horizontal axis. When its edge becomes exactly parallel to the slit, interference pattern consisting of alternate bright and dark bands can be seen through the eyepiece.

Measurements:

The wavelength of monochromatic light is given by the formula where

X = λD/d

Where d =  distance between two coherent sources.

D = distance between the slit and the focal plane of the eyepiece

Measurement of D:

D between the plane of the slit and the focal plane of the micrometer eyepiece can be directly measured from the scale marked at the edge of the optical bench.

For a given adjustment, reading corresponding to the reference marks on the bases of vertical stands carrying slit and eyepiece are noted.  The difference between the two readings gives D.

Measurement of X:

To measure fringe width X, the micrometer screw is so adjusted that, the cross wire of it coincides with one bright band. The micrometer reading x₁ is noted. By rotating the screw, the cross wire is made to coincide with successive bright bands and corresponding reading x₂, x₃, x₄, x₅, ….etc. are noted.  The average difference between two successive readings such as | x₂ – x₁|, | x₃ – x₂ |, |x₄ – x₃ |,……   etc. gives the mean fringe width.

Measurement of ‘d’ :

To measure ‘d’ a convex lens mounted on the stand kept between the biprism and the eyepiece, without disturbing the slit and the biprism, the eyepiece is moved in such a way that the distance between the slit and the eyepiece is more than four times the focal length of the lens. One lens is moved towards the slit and its position (L₁) is so adjusted that two magnified images of the slit are seen through the eyepiece.

By adjusting the eyepiece, its cross wire is made to coincide with each image and corresponding micrometer readings are noted.  The difference between the readings gives the distance d between the two magnified images.

By the principle of linear magnification

d₁/d = v /u    …….. (1)

The convex lens is now moved towards the eyepiece and its position (L₂) is adjusted so that two diminished images of the slit are seen through the eyepiece.

By adjusting the eyepiece, its cross wire is made to coincide with each image and corresponding micrometer readings are noted.  The difference between the readings gives the distance ‘d’ between the two diminished images.

By the principle of conjugate focii

d₂/d = u /v    …….. (2)

The wavelength of monochromatic light can be calculated using the formula X = λD/d

Previous Topic: Numerical Problems on Formation of Bright and dark bands

Next Topic: Numerical Problems on Fringe Width and Change of Fringe Width

Science > Physics > Interference of Light > Concept of Fringe Width and Path Difference

The post Concept of Fringe Width and Path Difference appeared first on The Fact Factor.

In this article we shall study problems, to determine the nature of interference at a point i.e. to decide whether the point is a bright point or dark point.

Example – 01:

The optical path difference between two identical waves arriving at a point is 371λ. Is the point bright or dark? If the path difference is 0.24 mm, calculate the wavelength of light used.

Given: Path difference = 371λ, path difference = 0.24 mm = 0.24 x 10^-3 m = 2.4 x 10^-4 m

To Find: Nature of illumination and wavelength λ =?

Solution:

Path difference = 371λ = 742 (λ/2)

Thus the path difference is even multiple of (λ/2)

Thus constructive interference takes place at the point. Hence the point is the bright point.

Given 371λ = 2.4 x 10^-4

∴ λ = (2.4/371) x 10^-4 = 6.469 x 10^-7 m

∴ λ = 6469 x 10^-10 m = 6469 Å

Ans: The point is a bright point and wavelength of light used is 6469 Å

Example – 02:

The optical path difference between two identical waves arriving at a point is 93 wavelengths. Is the point bright or dark? If the path difference is 46.5 micron, calculate the wavelength of light used.

Given: Path difference = 371λ, path difference = 0.24 mm = 46.5 micron = 46.5 x 10^-6 m

To Find: Nature of illumination and wavelength λ =?

Solution:

Path difference = 93λ = 186 (λ/2)

Thus the path difference is even multiple of (λ/2)

Thus constructive interference takes place at the point. Hence the point is the bright point.

Given 93λ = 46.5 x 10^-6

∴ λ = (46.5/93) x 10^-6 = 5 x 10^-7 m

∴ λ = 5000 x 10^-10 m = 5000 Å

Ans: The point is a bright point and wavelength of light used is 5000 Å

Example – 03:

The optical path difference between two identical waves arriving at a point is 85.5λ. Is the point bright or dark? If the path difference is 42.5 µm, calculate the wavelength of light used.

Given: Path difference = 85.5λ, path difference = 42.5 µm = 42.5 x 10^-6 m

To Find: Nature of illumination and wavelength λ =?

Solution:

Path difference = 85.5λ = 171 (λ/2)

Thus the path difference is the odd multiple of (λ/2)

Thus destructive interference takes place at the point. Hence the point is a dark point.

Given 85.5 λ = 42.5 x 10^-6

∴ λ = (42.5/85.5) x 10^-6 = 4.970 x 10^-7 m

∴ λ = 4970 x 10^-10 m = 4970 Å

Ans: The point is a dark point and wavelength of light used is 4970 Å

Example – 04:

The optical path difference between two identical waves arriving at a point is 6 x 10^-6 m. The wavelength of light used is 5000 Å. Is the point bright or dark? What is the number of the band?

Given: Path difference = 6 x 10^-6 m, Wavelength of light used = 5000 Å = 5000 x 10^-10 m = 5 x 10^-7 m

To Find: Nature of illumination and wavelength λ =?

Solution:

Path difference = kλ = 6 x 10^-6

∴ k x 5 x 10^-7 = 6 x 10^-6

∴ k = (6 x 10^-6) / (5 x 10^-7)= 12

∴ Path difference = 12λ = 24(λ /2)

Thus the path difference is even multiple of (λ/2)

Thus constructive interference takes place at the point. Hence the point is 12^th bright point.

Ans: The point is 12^th bright point

Example – 05:

The optical path difference between two identical waves arriving at a point is 185.5λ. Is the point bright or dark? If the path difference is 8.348 x 10^-3 cm, calculate the wavelength of light used.

Given: Path difference = 85.5λ, path difference = 8.348 x 10^-3 cm = 8.348 x 10^-5 m

To Find: Nature of illumination and wavelength λ =?

Solution:

Path difference = 185.5λ = 371 (λ/2)

Thus the path difference is an odd multiple of (λ/2)

Thus destructive interference takes place at the point

Hence the point is a dark point.

Given 185.5 λ = 8.348 x 10^-5

∴ λ = (8.348/185.5) x 10^-5 = 4.5 x 10^-7 m

∴ λ = 4500 x 10^-10 m = 4500 Å

Ans: The point is a dark point and wavelength of light used is 4500 Å

Example – 06:

The optical path difference between two identical waves arriving at a point is 13λ. Is the point bright or dark? If the path difference is 0.0078 mm, calculate the wavelength of light used.

Given: Path difference = 13λ, path difference = 0.0078 mm = 0.0078 x 10^-3 m = 7.8 x 10^-6 m

To Find: Nature of illumination and wavelength λ =?

Solution:

Path difference = 13λ = 26 (λ/2)

Thus the path difference is even multiple of (λ/2)

Thus constructive interference takes place at the point. Hence the point is a bright point.

Given 13λ = 7.8 x 10^-6

∴ λ = (7.8/13) x 10^-4 = 6 x 10^-7 m

∴ λ = 6000 x 10^-10 m = 6000 Å

Ans: The point is a bright point and wavelength of light used is 6000 Å

Example – 07:

The optical path difference between two identical waves arriving at a point is 165.5λ. Is the point bright or dark? If the path difference is 9.75 x 10^-5 m, calculate the wavelength of light used.

Given: Path difference = 371λ, path difference = 9.75 x 10^-3 m

To Find: Nature of illumination and wavelength λ =?

Solution:

Path difference = 165.5λ = 331 (λ/2)

Thus the path difference is an odd multiple of (λ/2)

Thus destructive interference takes place at the point. Hence the point is a dark point.

Given 165.5λ = 9.78 x 10^-5

∴ λ = (9.78 /165.5) x 10^-5 = 5.891 x 10^-7 m

∴ λ = 5891 x 10^-10 m = 5891 Å

Ans: The point is a dark point and wavelength of light used is 5891 Å

Example – 08:

The slits in the interference experiment are illuminated by the light of wavelength 5600 Å. Find the path difference and a phase difference of the waves arriving at a point P on the screen to form 8^th dark fringe.

Given: Wavelength of light used = 5600 Å = 5600 x 10^-10 m = 5.6 x 10^-7 m, 8^th dark band, n = 8

To Find: Path difference and phase difference =?

Solution:

For dark band the path difference = (2n – 1) λ/2

∴ Path difference = (2 x 8 – 1)λ/2 = 7.5 λ = 7.5 x 5.6 x 10^-7 = 4.2 x 10^-6 m

Thus the phase difference = (2n – 1) π = (2 x 8 – 1)π = 15π

Ans: The path difference is 4.2 x 10^-6 m and phase difference is 15π

Example – 09:

A point is situated at 6.5 cm and 6.65 cm from two coherent sources. Find the nature of the illumination of the point if wavelength of light is 5000 Å

Given: Distance of points from sources = 6.5 cm and 6.65 cm, wavelength of light used = λ = 5000 Å = 5000 x 10^-10 m = 5 x 10^-7 m

To Find: Nature of illumination =?

Solution:

Path difference = 6.65 cm – 6.5 cm = 0.15 cm = 0.15 x 10^-2 m = 1.5 x 10^-3 m

∴ Path difference = kλ = 1.5 x 10^-3

∴ k x 5 x 10^-7 = 1.5 x 10^-3

∴ k = (1.5 x 10^-3) / (5 x 10^-7)= 3000

∴ Path difference = 3000λ = 6000(λ /2)

Thus the path difference is even multiple of (λ/2), Thus constructive interference takes place at the point. Hence the point is a bright point.

Ans: The point is a bright point

Example – 10:

A point P is situated at 8.7 cm and 8.8 cm from two coherent light sources. Is the point bright or dark? l = 5000 A.U.

Given: Distance of points from sources = 8.7 cm and 8.8 cm, wavelength of light used = λ = 5000 Å = 5000 x 10^-10 m = 5 x 10^-7 m

To Find: Nature of illumination =?

Solution:

Path difference = 8.8 cm – 8.7 cm = 0.1 cm = 0.1 x 10^-2 m = 1 x 10^-3 m

Path difference = kλ = 1 x 10^-3

∴ k x 5 x 10^-7 = 1 x 10^-3

∴ k = (1.5 x 10^-3) / ( 5 x 10^-7)= 2000

∴ Path difference = 2000λ = 4000(λ /2)

Thus the path difference is even multiple of (λ/2), Thus constructive interference takes place at the point. Hence the point is a bright point.

Ans: The point is a bright point

Example – 11:

A screen is placed at a distance from two images of an illuminated slit. Distances of a point on the screen from the two images are 1.8 x 10^-5 m and 1.23 x 10^-5 m. If the wavelength of light used is 6000 A.U., is the point bright or dark? What is the number of bright or dark fringe formed at the point?

Given: Distance of points from sources = 1.8 x 10^-5 m and 1.23 x 10^-5 m, wavelength of light used = λ = 6000 Å = 6000 x 10^-10 m = 6 x 10^-7 m

To Find: Nature of illumination =?

Solution:

Path difference = 1.8 x 10^-5 – 1.23 x 10^-5 m = 0.57 x 10^-5 m = 5.7 x 10^-6 m

Path difference = kλ = 5.7 x 10^-6

∴ k x 6 x 10^-7 = 5.7 x 10^-6

∴ K = (5.7 x 10^-6)/( 6 x 10^-7)= 9.5

∴ Path difference = 9.5λ = 19(λ /2)

∴ 2n – 1 = 19

∴ 2n = 20

∴ n = 10

Thus the path difference is even multiple of (λ/2), Thus constructive interference takes place at the point. Hence the point is a 10^th dark point.

Ans: The point is a 10^th bright point

Example – 12:

In Young’s experiment, the wavelength of monochromatic light used is 6000 Å. The optical path difference between the rays from the two coherent sources at point P on the screen is 0.0075 mm and at point Q on the screen is 0.0015 mm. How many bright and dark bands are observed between the two points P and Q. (Points P and Q are on the opposite sides of the central bright band)

Given: Wavelength of light used = λ = 6000 Å = 6000 x 10^-10 m = 6 x 10^-7 m

To Find: Number of bright and dark bands between P and Q.

Solution:

Consider point P:

Path difference = 0.0075 mm = 0.0075 x 10^-3 m = 7.5 x 10^-6 m

Path difference = kλ = 7.5 x 10^-6

∴ k x 6 x 10^-7 = 7.5 x 10^-6

∴ k = (7.5 x 10^-6) / (6 x 10^-7)= 12.5

∴ Path difference = 12.5λ = 25(λ /2)

∴ (2n – 1) = 25

∴ 2n = 26

∴ n = 13

Thus the path difference is odd multiple of (λ/2). Thus destructive interference takes place at the point. Hence the point P is 13^th dark point.

Consider point Q:

Path difference = 0.0015 mm = 0.0015 x 10^-3 m = 1.5 x 10^-6 m

Path difference = kλ = 1.5 x 10^-6

∴ k x 6 x 10^-7 = 1.5 x 10^-6

∴ k = (1.5 x 10^-6) / (6 x 10^-7)= 2.5

∴ Path difference = 2.5λ = 5(λ /2)

∴ (2n – 1) = 5

∴ 2n = 6

∴ n = 3

Thus the path difference is an odd multiple of (λ/2). Thus destructive interference takes place at the point. Hence the point P is 3^rd dark point.

Points P and Q are on the opposite sides of the central bright band

Hence the number of dark bands between P and Q

= 13 + 3 = 16 (including those at P and Q) Or 14 (excluding those at P and Q)

Hence the number of bright bands between P and Q

= (13 -1) + (3 – 1) + 1 central band = 12 + 2 + 1 = 15

Ans: Number of bright bands = 15 and number of dark bands = 14

Example – 13:

In Young’s experiment, the wavelength of monochromatic light used is 4000 Å. The optical path difference between the rays from the two coherent sources at point P on the screen is 0.0080 mm and at point Q on the screen is 0.0022 mm. How many bright and dark bands are observed between the two points P and Q. (Points P and Q are on the opposite sides of the central bright band)

Given: Wavelength of light used = λ = 4000 Å = 4000 x 10^-10 m = 6 x 10^-7 m

To Find: Number of bright and dark bands between P and Q.

Solution:

Consider point P:

Path difference = 0.0080 mm = 0.0080 x 10^-3 m = 8 x 10^-6 m

Path difference = kλ = 8 x 10^-6

∴ k x 4 x 10^-7 = 8 x 10^-6

∴ k = (8 x 10^-6) / (4 x 10^-7) = 20

Path difference = 20λ = 40(λ /2)

n = 20

Thus the path difference is even multiple of (λ/2). Thus constructive interference takes place at the point. Hence the point P is 20^th bright point.

Consider point Q:

Path difference = 0.0022 mm = 0.0022 x 10^-3 m = 2.2 x 10^-6 m

Path difference = kλ = 2.2 x 10^-6

∴ k x 4 x 10^-7 = 2.2 x 10^-6

∴ k = (2.2 x 10^-6) / (4 x 10^-7)= 5.5

∴ Path difference = 5.5λ = 11(λ /2)

∴ (2n – 1) = 11

∴ 2n = 12

∴ n = 6

Thus the path difference is an odd multiple of (λ/2). Thus destructive interference takes place at the point. Hence the point P is 6^th dark point.

Points P and Q are on the opposite sides of the central bright band

Hence the number of bright bands between P and Q

= 20 + 1 central band + (6 – 1) = 21 + 5 = 26 (including that at P) Or 25 (excluding that at P)

The number of dark bands between P and Q

= 20 + 6 = 26 (including that at Q) 0r 25 (excluding that at Q)

Ans: Number of bright bands = 25 and number of dark bands = 25

Example – 14:

A phase difference between the coherent waves of wavelength λ originating from two slits is 3π radian. A point P on the screen is at a distance of 20λ and 21.5λ from the two slits. Is point P bright or dark?

To Find: Nature of illumination =?

Solution:

Path difference = 21.5λ – 20λ = 1.5λ

Phase difference of source = 3π radian = (3/2)(2π) radian

2π radian corresponds to λ

Hence 3π radian corresponds to 1.5λ

Thus the phase difference of waves meeting at P is 1.5λ ± 1.5λ = 3λ or 0

In either case, it is an integral multiple of λ. Hence point P is the bright point.

Ans: Point P is a bright point

Previous Topic: Interference of Light

Next Topic: Concept of Fringe Width and Path Difference

Science > Physics > Interference of Light > Numerical Problems on Bright and Dark Bands

The post Numerical Problems on Formation of Bright and Dark Band appeared first on The Fact Factor.

In this article, we shall study the phenomenon of interference of light, conditions for constructive and destructive interference.

Terminology:

Constructive Interference of Light:

When two light waves of same frequency arriving at a point, meet each other in the same phase i.e. the crest due to one wave matches with the crest due to other wave and the trough due to first wave matches with the trough due to another wave, then the interference is called constructive interference.

Destructive Interference of Light:

When two light waves of same frequency arriving at a point, meet each other in opposite phase i.e. the crest due to one wave matches with the trough due to other wave and the trough due to first wave matches with the crest due to another wave, then the interference is called destructive interference.

Interference of Light:

The modification in the intensity of light (redistribution of light energy) produced by the superposition of two or more light waves is called interference of light.

Steady Interference Pattern:

An interference pattern in which the intensity of light at any given point remains constant is called a steady or stationary interference pattern.

Monochromatic Sources of Light:

Two sources of light are said to be monochromatic only when they emit light of the same wavelength, i.e. same frequency.

Coherent Sources:

Two sources are said to be coherent if there always exists a constant phase difference between the waves emitted by these sources.

Principle of Superposition of Waves:

Statement: If two or more waves arrive at a point simultaneously, each wave produces its own displacement (effect) at that point as if the other waves are not present.

Hence the resultant displacement of that point at that instant is the vector sum of the displacements due to all the waves.

If two waves arrive at a point simultaneously, and the displacements due to these waves at that point are y₁ and y₂ respectively, then by the principle of superposition of wave resultant displacement at that point is given by  

y = y₁   + y₂

The Phenomenon of Interference of Light:

The phenomenon of enhancement or cancellation of displacements and the subsequent redistribution of intensity is called interference. The modification in the intensity of light (redistribution of light energy) produced by the superposition of two or more light waves is called interference of light.

When two light waves of same frequency arriving at a point, meet each other in the same phase i.e. the crest due to one wave matches with the crest due to other wave and the trough due to first wave matches with the trough due to another wave, then the constructive interference takes place. At this point intensity is maximum and maximum intensity or brightness is obtained at that point. This point is the bright point.

When two light waves of same frequency arriving at a point, meet each other in opposite phase i.e. the crest due to one wave matches with the trough due to other wave and the trough due to first wave matches with the crest due to another wave, then the destructive interference takes place. At this point intensity is the minimum and minimum intensity or brightness is obtained at that point. This point is the dark point.

If this change in intensity is of light due to interference is obtained on screen, then alternate bright and dark bands or rings are obtained. This pattern is known as the interference pattern.

Conditions for Constructive Interference of Light (Formation of Bright Point):

In order to obtain constructive interference or brightness at a point, the two light waves from the two sources should arrive at the point in the same phase.

Thus for brightness, required path difference is 0, λ, 2λ, 3λ, 4λ, 5λ, …. etc. or phase difference is 0, 2π, 4π, 6π, 8π,… etc.

In general for obtaining a bright band at a point, Path difference = nλ where n = 0, 1,2,3…….

Conditions for Destructive Interference of Light (Formation of dark Point):

In order to obtain destructive interference or darkness at the point, the two light waves from the two sources should arrive at the point in the opposite phase.

Thus for darkness, the required path difference is λ/2, 3λ/2, 5λ/2, …. etc. or phase difference is π, 3π, 5π, 7π,… etc.

In general for obtaining a dark band at a point, Path difference = (2n – 1)λ/2 where n = 1,2,3……. etc.

Steady Interference Pattern:

An interference pattern in which the intensity of light at any given point remains constant is called a steady or stationary interference pattern. In such a pattern intensity of light not changes with time i.e. bright point remains bright and a dark point always remains dark. The steady interference pattern consists of alternately bright and dark bands or rings called interference fringes.

Conditions for Obtaining Steady Interference Pattern:

• The two sources of light must be coherent. That is these two sources should emit light waves with a constant phase difference. In such a case, coherent sources are obtained from a single source.
• The amplitudes of waves must be equal. i.e. the brightness of the two sources should be the same.
• The sources of light should be narrow.
• The two sources of light must be monochromatic.
• The distance between the sources and the screen must be reasonably large.
• The distance between coherent sources must be small.
• The sources should emit light waves in nearly the same direction.
• The two interfering waves must be in the same state of polarization.

For obtaining steady interference pattern the two sources of light must be coherent:

Two sources are said to be coherent if there always exists a constant phase difference between the waves emitted by these sources. Two sources are coherent only if they are obtained from one and the same original sources e.g. the two images produced by a biprism of a monochromatic slit can be treated as coherent sources.

It is essential to have coherent sources to produce a stationary interference pattern. Because, if the sources are not coherent then the phase difference will change abruptly at any point and hence the intensity of light will change abruptly.  But when the sources are coherent, then the resultant intensity of light at a point will remain constant and so interference fringes will remain stationary.

For obtaining steady interference pattern the light used is monochromatic:

Two sources of light are said to be monochromatic only when they emit light of the same wavelength and the same frequency. If the sources emit light waves of more than one wavelength (frequency) then constructive Interference or brightness at a point due to one may overlap with destructive Interference or darkness due to the other wavelength.  Thus interference pattern will be diffused. Therefore, the two coherent sources of light should be monochromatic.

For obtaining steady interference pattern the two light sources   should be of equal intensity:

The intensity of light is directly proportional to the square of the amplitude of light waves. In an interference pattern, at some points, complete darkness or zero intensity can be obtained only when the interfering waves arrive at that point in exactly the opposite phase and cancel each other. Therefore, the two sources should emit light waves of same amplitude i.e. same intensity (The sources should be equally bright)

For obtaining steady interference pattern the two sources of light should be narrow:

A wide source of light is equivalent to a number of narrow sources. If the sources are broad, then light waves emitted from different points of the same source itself would interfere.  This, therefore causes some pattern to overlap above the actual interference bands making it diffused or less sharp.

For obtaining steady interference pattern the two sources of light should be close to each other:

In an interference pattern, the fringe width or bandwidth (X) is inversely proportional to the distance between the two coherent sources (d). When these sources are close to each other, the fringes will be widely spaced and can be seen clearly. This helps in measuring fringe width accurately.

For obtaining steady interference pattern the distance between the sources and the screen should be large:

In an interference pattern, the fringe width or bandwidth (X) is directly proportional to the distance between the sources and the screen (D). When this distance is large, the fringes will be widely spaced and can be seen clearly. This helps in measuring fringe width accurately.

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