Home
Class 12
PHYSICS
In a Young's double slit experiment, (sl...

In a Young's double slit experiment, (slit distance d) monochromatic light of wavelength `lambda` is used and the fringe pattern observed at a distance D from the slits. The angular position of the bright fringes are

A

` "sin"^(-1)((N lambda)/(d))`

B

` "sin"^(-1)(((N+(1)/(2)) lambda)/(d))`

C

` "sin"^(-1)((N lambda)/(D))`

D

` "sin"^(-1)(((N+(1)/(2)) lambda)/(D))`

Text Solution

AI Generated Solution

The correct Answer is:
To find the angular position of the bright fringes in a Young's double slit experiment, we can follow these steps: ### Step-by-Step Solution: 1. **Understand the Setup**: In a Young's double slit experiment, we have two slits separated by a distance \(d\) and a screen placed at a distance \(D\) from the slits. Monochromatic light of wavelength \(\lambda\) is used. 2. **Identify the Path Difference**: The path difference \(\Delta x\) between the light waves coming from the two slits to a point on the screen is given by: \[ \Delta x = \frac{d \cdot y}{D} \] where \(y\) is the distance from the central maximum to the point of interest on the screen. 3. **Condition for Bright Fringes**: For constructive interference (bright fringes), the path difference must be an integer multiple of the wavelength: \[ \Delta x = n\lambda \] where \(n\) is an integer (0, 1, 2, ...). 4. **Set Up the Equation**: Equating the two expressions for path difference: \[ \frac{d \cdot y}{D} = n\lambda \] 5. **Rearrange to Find \(y\)**: Rearranging gives: \[ y = \frac{n\lambda D}{d} \] 6. **Find the Angular Position**: The angular position \(\theta\) of the bright fringe can be found using the small angle approximation: \[ \sin \theta \approx \tan \theta = \frac{y}{D} \] Substituting for \(y\): \[ \sin \theta \approx \frac{n\lambda D}{dD} = \frac{n\lambda}{d} \] 7. **Final Expression for Angular Position**: Thus, the angular position of the bright fringes is given by: \[ \theta = \sin^{-1}\left(\frac{n\lambda}{d}\right) \] ### Summary: The angular position of the bright fringes in a Young's double slit experiment is: \[ \theta = \sin^{-1}\left(\frac{n\lambda}{d}\right) \]

To find the angular position of the bright fringes in a Young's double slit experiment, we can follow these steps: ### Step-by-Step Solution: 1. **Understand the Setup**: In a Young's double slit experiment, we have two slits separated by a distance \(d\) and a screen placed at a distance \(D\) from the slits. Monochromatic light of wavelength \(\lambda\) is used. 2. **Identify the Path Difference**: ...
Promotional Banner

Topper's Solved these Questions

  • WAVE OPTICS

    NCERT FINGERTIPS ENGLISH|Exercise HIGHER ORDER THINKING SKILLS|8 Videos
  • WAVE OPTICS

    NCERT FINGERTIPS ENGLISH|Exercise NCERT EXEMPLAR PROBLEMS|5 Videos
  • SEMICONDUCTOR ELECTRONICS : MATERIALS , DEVICES AND SIMPLE CIRCUITS

    NCERT FINGERTIPS ENGLISH|Exercise Assertion And Reason|15 Videos

Similar Questions

Explore conceptually related problems

In Yonung's double-slit experiment, two slits which are separated by 1.2 mm are illuminated with a monochromatic light of wavelength 6000 Å The interference pattern is observed on a screen placed at a distance of 1 m from the slits. Find the number of bright fringes formed over 1 cm width on the screen.

In Young's double slit experiment , the two slits 0.20 mm apart are illuminated by monochromatic light of wavelength 600 nm . The screen 1.0 m away from the slits . (a) Find the distance of the second (i) bright fringe , (ii) dark fringe from the central maximum . (b) How will the fringe pattern change if the screen is moved away from the slits ?

In Young's double slit experiment the two slits 0.12 mm apart are illuminated by monochromatic light of wavelength 420 nm. The screen is 1.0 m away from the slits . (a) Find the distance of the second (i) bright fringe , (ii) dark fringe from the central maximum . (b) How will the fringe pattern change if the screen is moved away from the slits ?

In Young.s double slit experiment using monochromatic light of wavelength 600 nm, 5th bright fringe is at a distance of 0.48 mm from the centre of the pattern. If the screen is at a distance of 80 cm from the plane of the two slits, calculate : Fringe width, i.e., fringe separation.

In Young.s double slit experiment using monochromatic light of wavelength 600 nm, 5th bright fringe is at a distance of 0.48 mm from the centre of the pattern. If the screen is at a distance of 80 cm from the plane of the two slits, calculate : Distance between the two slits.

Find the angular fringe width in Young's double slit experiment using a monochromatic light of wavelength 500 nm when the slits are separated by 2.5 mm.

Statement-1: In Young's double slit experiment the two slits are at distance d apart. Interference pattern is observed on a screen at distance D from the slits. At a poit on the screen when it is directly opposite to one of the slits, a dard fringe is observed then the wavelength of wave is proportional of square of distance of two slits. Statement-2: In Young's double slit experiment, for identical slits, the intensity of a dark fringe is zero.

In a Young's double slit experiment, light of 500 nm is used to produce an interference pattern. When the distance between the slits is 0.05 mm, the angular width (in degree) of the fringes formed on the distance screen is close to:

In Young's double slit experiment, the distance between two slits is made three times then the fringe width will becomes

Statement I: In Young's double-slit experiment, the two slits are at distance d apart. Interference pattern is observed on a screen at distance D from the slits. At a point on the screen which is directly opposite to one of the slits, a dark fringe is observed. Then, the wavelength of wave is proportional to the squar of distance between two slits. Statement II: For a dark fringe, intensity is zero

NCERT FINGERTIPS ENGLISH-WAVE OPTICS-Assertion And Reason
  1. In a Young's double slit experiment, (slit distance d) monochromatic l...

    Text Solution

    |

  2. Assertion : The frequencies of incident, reflected and refracted beam ...

    Text Solution

    |

  3. Assertion: When a light wave travels from a rarer to a denser medium, ...

    Text Solution

    |

  4. Assertion : Wavefronts obtained from light emitted by a point source i...

    Text Solution

    |

  5. Assertion : When a plane wave passes through a thin prism, the emergin...

    Text Solution

    |

  6. Assertion : The increase in wavelength due to doppler effect is termed...

    Text Solution

    |

  7. Assertion : Interference is not observed if the two coherent slit sour...

    Text Solution

    |

  8. Assertion : When a thin transparent sheet is placed in front of both t...

    Text Solution

    |

  9. Statement-I : In Young's double slit experiment interference pattern d...

    Text Solution

    |

  10. Assertion : The fringe closest on either side of the central white fri...

    Text Solution

    |

  11. Assertion : All bright interference bands have same intensity. Reaso...

    Text Solution

    |

  12. Assertion : If we look clearly at the shadow cast by an opaque object,...

    Text Solution

    |

  13. Assertion : If the light from an ordinary source passes through a pola...

    Text Solution

    |

  14. Assertion : Sound waves cannot be polarised. Reason : Sound waves ar...

    Text Solution

    |

  15. Assertion : In interference and diffraction, light energy is redistrib...

    Text Solution

    |

  16. Assertion : Intensity pattern of interference and diffraction are not ...

    Text Solution

    |