Two slits in Young's experiment are 0 . 02 cm apart. The interference fringes for the light of wavelength 6000 A ∘ are formed on a screen 80 cm away. Calculate the distance of the fifth bright fringe.

Given, D = 80 cm = 80 × 10 - 2 m d = 0 . 02 cm = 0 . 02 × 10 - 2 m Now distance of fifth bright fringe = 5 y where y = fringe width = D λ d = 5 × D λ d = 5 × 80 × 10 - 2 × 6000 × 10 - 10 0 . 02 × 10 - 2 = 12 × 10 - 3 m = 12 mm = 1 . 2 cm This is the required answer.

Two slits in Young's experiment are 0 . 02 cm apart. The interference fringes for the light of wavelength 6000 A ∘ are formed on a screen 80 cm away. Calculate the distance of the fifth bright fringe.

Given, D = 80 cm = 80 × 10 - 2 m d = 0 . 02 cm = 0 . 02 × 10 - 2 m Now distance of fifth bright fringe = 5 y where y = fringe width = D λ d = 5 × D λ d = 5 × 80 × 10 - 2 × 6000 × 10 - 10 0 . 02 × 10 - 2 = 12 × 10 - 3 m = 12 mm = 1 . 2 cm This is the required answer.

In a two slits experiment with monochromatic light, fringes are obtained on a screen placed at some distance from the slits. If the screen is moved by 5 × 10 - 2 m towards the slits, the change in the fringe width is 3 × 10 - 5 m . If the distance between the slits is 10 - 3 m , calculate the wavelength of light used

The formula used β = λ D d where β is fringe width λ is the wavelength of light D is the distance between slit and screen. d is the distance between slits. Change in fringe width when screen distance from slits is changed by ∆ D is given by ⇒ ∆ β = λ ( ∆ D ) d ⇒ λ = d ( ∆ β ) ∆ D = ( 10 - 3 ) ( 3 × 10 - 5 ) 5 × 10 - 2 ⇒ λ = 6000 A o

In young's experiment for interference of light the slits 0 . 2 cm apart are illuminated by yellow light λ = 5896 A ∘ . What would be the fringe width on a screen placed 1 m from the plane of slits. What will be the fringe width, if the system is immersed in water. μ w = 4 3 ?

Given that the slits are separated by a distance of d = 0 . 2 cm = 2 × 10 - 3 m , Distance of the screen from the plane of slits is D = 1 m , Wavelength of the yellow light in air is λ y a = 5896 A o . So, the fringe width β can be given as; β = λ y a D d = 5896 × 10 - 10 × 1 2 × 10 - 3 = 2948 × 10 - 7 m = 0 . 2948 mm . Now, when the system is immersed in water then, the wavelength of the light in water would be given as; λ y w = λ y a μ = 5896 4 3 = 5896 × 3 4 A o , So, the fringe width can be given as; β w = λ y w D d = 5896 × 10 - 10 × 3 × 1 4 × 2 × 10 - 3 = 0 . 2211 mm . Hence, the required fringe widths are 0 . 2948 mm and 0 . 2211 mm .

In Young's experiment, the width of the fringes obtained with a light of wavelength 6000 A ∘ is 2 . 0 mm . What will be the fringe width, if the entire apparatus is immersed in a liquid of refractive index 1 . 33 ?

Given, Width of fringes, ω = 2 × 10 - 3 m Wavelength of light used, λ = 6000 A ∘ = 6 × 10 - 7 The formula for fringe width is, ω = λ D d D d = ω λ D d = 2 × 10 - 3 6 × 10 - 7 = 1 3 × 10 4 When the apparatus is immersed in liquid Wavelength, λ ' = λ μ = 6 × 10 - 7 1 . 33 m Fringe width, ω ' = D d λ ' ω ' = 1 3 × 10 4 × 6 × 10 - 7 1 . 33 m = 1 . 5 × 10 - 3 m = 1 . 5 mm This is the required answer.

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Earn 100

Two slits in Young's experiment are $0.02 cm$ apart. The interference fringes for the light of wavelength $6000 \overset{\circ}{A}$ are formed on a screen $80 cm$ away. Calculate the distance of the fifth bright fringe.

Important Questions on Wave Optics

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Beta Question Bank for Medical: Physics>Chapter 31 - Wave Optics>BEGINNER'S BOX - 2>Q 4

In Young's double-slit experiment, two slits are separated by

3 m m

distance and illuminated by light of wavelength

480 n m

. The screen is at

2 m

from the plane of the slits. Calculate the separation between the

8^{t h}

bright fringe and the

3^{r d}

dark fringe observed with respect to the central bright fringe.

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Beta Question Bank for Medical: Physics>Chapter 31 - Wave Optics>BEGINNER'S BOX - 2>Q 5

In a two slits experiment with monochromatic light, fringes are obtained on a screen placed at some distance from the slits. If the screen is moved by

5 \times 10^{- 2} m

towards the slits, the change in the fringe width is

3 \times 10^{- 5} m

. If the distance between the slits is

10^{- 3} m

, calculate the wavelength of light used

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NEET

IMPORTANT

Beta Question Bank for Medical: Physics>Chapter 31 - Wave Optics>BEGINNER'S BOX - 2>Q 6

In young's experiment for interference of light the slits

0.2 cm

apart are illuminated by yellow light

(λ = 5896 \overset{\circ}{A})

. What would be the fringe width on a screen placed

1 m

from the plane of slits. What will be the fringe width, if the system is immersed in water.

(μ_{w} = \frac{4}{3})

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Beta Question Bank for Medical: Physics>Chapter 31 - Wave Optics>BEGINNER'S BOX - 2>Q 7

The distance between the coherent source is

0.3 mm

and the screen is

90 cm

from the sources. The second dark band is

0.3 cm

away from the central bright fringe. Find the wavelength and the distance of the fourth bright fringe from the central bright fringe.

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Beta Question Bank for Medical: Physics>Chapter 31 - Wave Optics>BEGINNER'S BOX - 2>Q 8

State two conditions to obtain sustained interference of light. In Young's double-slit experiment, using light of wavelength

400 nm

, interference fringes of width

X

are obtained. The wavelength of light is increased to

600 nm

and the separation between the slits is halved. If one wants the observed fringe width on the screen to be the same in the two cases, find the ratio of the distance between the screen and the plane of the slits in the two arrangements.

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Beta Question Bank for Medical: Physics>Chapter 31 - Wave Optics>BEGINNER'S BOX - 2>Q 9

In a Young's double slit experiment, light has a frequency of

6 \times 10^{14} s^{- 1}

. The distance between the centres of adjacent bright fringes is

0.75 m m

. If the screen is

1.5 m

away, what is the distance between the slits?

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Beta Question Bank for Medical: Physics>Chapter 31 - Wave Optics>BEGINNER'S BOX - 2>Q 10

In Young's experiment, the width of the fringes obtained with a light of wavelength

6000 \overset{\circ}{A}

2.0 mm

. What will be the fringe width, if the entire apparatus is immersed in a liquid of refractive index

1.33

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Beta Question Bank for Medical: Physics>Chapter 31 - Wave Optics>BEGINNER'S BOX - 2>Q 11

The intensity of X-rays of wavelength

0.5 \overset{°}{A}

reduces to one-fourth on passing through

3.5 mm

thickness of a metal foil. The coefficient of absorption of metal will be:-

Two slits in Young's experiment are 0.02 cm apart. The interference fringes for the light of wavelength 6000 A∘ are formed on a screen 80 cm away. Calculate the distance of the fifth bright fringe.

Important Questions on Wave Optics

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Two slits in Young's experiment are $0.02 cm$ apart. The interference fringes for the light of wavelength $6000 \overset{\circ}{A}$ are formed on a screen $80 cm$ away. Calculate the distance of the fifth bright fringe.