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The maximum intensity in Young's double ...

The maximum intensity in Young's double slit experiment is `I_(0)`. Distance between the slits is `d=2lamda`, where `lamda` is the wavelength of monochromatic light used in experiment. What will be the intensity of light in front of one of the slits on a screen at a distance `D=6d`?

A

`(I_(0))/2`

B

`3/4I_(0)`

C

`I_(0)`

D

`(I_(0))/4`

Text Solution

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The correct Answer is:
To solve the problem, we will follow these steps: ### Step 1: Understand the setup of the problem In Young's double slit experiment, we have two slits separated by a distance \( d \) and we are given that \( d = 2\lambda \), where \( \lambda \) is the wavelength of the light used. ### Step 2: Identify the distance to the screen The distance from the slits to the screen is given as \( D = 6d \). Since \( d = 2\lambda \), we can calculate \( D \): \[ D = 6d = 6 \times 2\lambda = 12\lambda \] ### Step 3: Determine the position in front of one of the slits The intensity at a point on the screen depends on the path difference between the light coming from the two slits. The point directly in front of one of the slits corresponds to a path difference of zero. ### Step 4: Calculate the path difference In this case, since we are looking at a point directly in front of one of the slits, the path difference \( \Delta x \) will be: \[ \Delta x = 0 \] ### Step 5: Calculate the phase difference The phase difference \( \Delta \phi \) is related to the path difference by the formula: \[ \Delta \phi = \frac{2\pi}{\lambda} \Delta x \] Since \( \Delta x = 0 \), we have: \[ \Delta \phi = 0 \] ### Step 6: Use the intensity formula The intensity \( I \) at any point on the screen in Young's double slit experiment can be expressed as: \[ I = I_0 \cos^2\left(\frac{\Delta \phi}{2}\right) \] Substituting \( \Delta \phi = 0 \): \[ I = I_0 \cos^2(0) = I_0 \times 1 = I_0 \] ### Step 7: Conclusion Thus, the intensity of light in front of one of the slits on the screen at a distance \( D = 6d \) is: \[ I = I_0 \] ### Final Answer: The intensity of light in front of one of the slits on the screen is \( I_0 \). ---

To solve the problem, we will follow these steps: ### Step 1: Understand the setup of the problem In Young's double slit experiment, we have two slits separated by a distance \( d \) and we are given that \( d = 2\lambda \), where \( \lambda \) is the wavelength of the light used. ### Step 2: Identify the distance to the screen The distance from the slits to the screen is given as \( D = 6d \). Since \( d = 2\lambda \), we can calculate \( D \): \[ ...
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The intensity at the maximum in a Young's double slit experiment is I_(0) . Distance between two slits is d=5 lambda , where lambda is the wavelength of light used in the experiment. The intensity infront of one of the slits on the screen placed at a distance D=10 d is (I_(0))/(n) . Find n .

Youngs Double slit experiment | variation of intensity on screen

Knowledge Check

  • The maximum intensity in young's double-slit experiment is I_(0) . Distance between the slit is d = 5 lambda , where lambda is the wavelength of monochromatic light used in the experiment. What will be the intensity of light in front of one of the slits on a screen at a distance D = 10 d ?

    A
    `I_(0)/(2)`
    B
    `(3)/(4) I_(0)`
    C
    `I_(0)`
    D
    `(I_(0))/(4)`
  • The maximum intensity in Young's double slit experiment is J_0 . Distance between the slits is d = 5lambda , where lambda is the wavelength of monochromatic light used in the experiment. What will be the intensity of light in front of one of the slits on a screen at a distance D = 10 d ?

    A
    `I_(0)/2`
    B
    `(3I_(0))/4`
    C
    `I_(0)`
    D
    `I_(0)/4`
  • The intensity at the maximum in Young,s double slit experiment is I_(0) . Distance between two slits is d = 5 lambda , where lambda is the wavelength of light used in the experiment. What will be the intensity in front one of the slits on the screen placed at a distance D = 10 d?

    A
    `I_(0)`
    B
    `(I_(0))/(4)`
    C
    `(3)/(4)I_(0)`
    D
    `(I_(0))/(2)`
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