हिंदी
कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान 2nd PUC Class 12

In a double-slit experiment using the light of wavelength 600 nm, the angular width of the fringe formed on a distant screen is 0.1°. Find the spacing between the two slits. - Physics

Advertisements
Advertisements

प्रश्न

In a double-slit experiment using the light of wavelength 600 nm, the angular width of the fringe formed on a distant screen is 0.1°. Find the spacing between the two slits.

संख्यात्मक
Advertisements

उत्तर १

The angular width (θ) of the fringe in the double-slit experiment is given by,

`theta=lambda/"d"`

Where

d = Spacing between the slits

Given:

The wavelength of light, λ = 600 nm

The angular width of the fringe,

θ = 0.1° = `pi/1800` = 0.0018 rad

∴ d = `lambda/theta`

d = `(600xx10^(-9))/(18xx10^(-4))`

d = 0.33 × 103 m

shaalaa.com

उत्तर २

Wavelength of light used, λ = 6000 nm = 600 × 10−9 m

Angular width of fringe, θ = 0.1° =` 0.1 xx pi/180 = 3.14/1800 "rad"`

The angular width of a fringe is related to slit spacing (d) as:

θ = `lambda/"d"`

`"d"= lambda/θ`

= `(600 xx 10^(-9))/(3.14 /1800)`

= 3.44 × 104 m

Therefore, the spacing between the slits is 3.44 × 104 m.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
2014-2015 (March) Delhi Set 2

संबंधित प्रश्न

The intensity at the central maxima in Young’s double slit experiment is I0. Find out the intensity at a point where the path difference is` lambda/6,lambda/4 and lambda/3.`


In a double-slit experiment the angular width of a fringe is found to be 0.2° on a screen placed 1 m away. The wavelength of light used is 600 nm. What will be the angular width of the fringe if the entire experimental apparatus is immersed in water? Take refractive index of water to be 4/3.


Explain two features to distinguish between the interference pattern in Young's double slit experiment with the diffraction pattern obtained due to a single slit.


In Young’s double slit experiment, show graphically how the intensity of light varies with distance


In Young’s experiment interference bands were produced on a screen placed at 150 cm from two slits, 0.15 mm apart and illuminated by the light of wavelength 6500 Å. Calculate the fringe width.


The slits in a Young's double slit experiment have equal width and the source is placed symmetrically with respect to the slits. The intensity at the central fringe is I0. If one of the slits is closed, the intensity at this point will be ____________ .


In a double slit interference experiment, the separation between the slits is 1.0 mm, the wavelength of light used is 5.0 × 10−7 m and the distance of the screen from the slits is 1.0m. (a) Find the distance of the centre of the first minimum from the centre of the central maximum. (b) How many bright fringes are formed in one centimetre width on the screen?


White light is used in a Young's double slit experiment. Find the minimum order of the violet fringe \[\left( \lambda = 400\text{ nm} \right)\] which overlaps with a red fringe \[\left( \lambda = 700\text{ nm} \right).\]


A parallel beam of monochromatic light is used in a Young's double slit experiment. The slits are separated by a distance d and the screen is placed parallel to the plane of the slits. Slow that if the incident beam makes an angle \[\theta =  \sin^{- 1}   \left( \frac{\lambda}{2d} \right)\] with the normal to the plane of the slits, there will be a dark fringe at the centre P0 of the pattern.


A double slit S1 − S2 is illuminated by a coherent light of wavelength \[\lambda.\] The slits are separated by a distance d. A plane mirror is placed in front of the double slit at a distance D1 from it and a screen ∑ is placed behind the double slit at a distance D2 from it (see the following figure). The screen ∑ receives only the light reflected by the mirror. Find the fringe-width of the interference pattern on the screen.


Consider the arrangement shown in the figure. By some mechanism, the separation between the slits S3 and S4 can be changed. The intensity is measured at the point P, which is at the common perpendicular bisector of S1S2 and S2S4. When \[z = \frac{D\lambda}{2d},\] the intensity measured at P is I. Find the intensity when z is equal to

(a) \[\frac{D\lambda}{d}\]

(b) \[\frac{3D\lambda}{2d}\]  and

(c) \[\frac{2D\lambda}{d}\]


A thin circular ring of mass M and radius R is rotating about its axis with a constant angular velocity ω. Two objects each of mass m are attached gently to the opposite ends of diameter of the ring. The ring will now rotate with an angular velocity:


The Young's double slit experiment is performed with blue and with green light of wavelengths 4360Å and 5460Å respectively. If x is the distance of 4th maxima from the central one, then:


ASSERTION (A): In an interference pattern observed in Young's double slit experiment, if the separation (d) between coherent sources as well as the distance (D) of the screen from the coherent sources both are reduced to 1/3rd, then new fringe width remains the same.

REASON (R): Fringe width is proportional to (d/D).


How will the interference pattern in Young's double-slit experiment be affected if the phase difference between the light waves emanating from the two slits S1 and S2 changes from 0 to π and remains constant?


The central fringe of the interference pattern produced by the light of wavelength 6000 Å is found to shift to the position of the fourth bright fringe after a glass plate of refractive index 1.5 is introduced in the path of one of the beams. The thickness of the glass plate would be ______.


Two beams of light having intensities I and 41 interfere to produce a fringe pattern on a screen. The phase difference between the two beams are π/2 and π/3 at points A and B respectively. The difference between the resultant intensities at the two points is xl. The value of x will be ______.


  • Assertion (A): In Young's double slit experiment all fringes are of equal width.
  • Reason (R): The fringe width depends upon the wavelength of light (λ) used, the distance of the screen from the plane of slits (D) and slits separation (d).

In Young's double slit experiment, show that:

`β = (λ"D")/"d"`

Where the terms have their usual meaning.


In Young’s double slit experiment, how is interference pattern affected when the following changes are made:

  1. Slits are brought closer to each other.
  2. Screen is moved away from the slits.
  3. Red coloured light is replaced with blue coloured light.

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×