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Science > Physics > Interference of Light > You are Here |

#### 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 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

Science > Physics > Interference of Light > You are Here |

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