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In Young's double slit experiment, we ge...

In Young's double slit experiment, we get `60` fringes in the field of view of monochromatic light of wavelength `4000Å`. If we use monochromatic light of wavelength `6000Å`, then the number of fringes obtained in the same field of view is

A

60

B

90

C

40

D

1.5

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To solve the problem, we will follow these steps: ### Step 1: Understand the relationship between the number of fringes and fringe width In Young's double slit experiment, the number of fringes (n) in a given field of view is given by the formula: \[ n = \frac{L}{\beta} \] where \( L \) is the length of the region and \( \beta \) is the fringe width. ### Step 2: Express the fringe width in terms of wavelength The fringe width \( \beta \) is given by: \[ \beta = \frac{\lambda D}{d} \] where \( \lambda \) is the wavelength of light, \( D \) is the distance from the slits to the screen, and \( d \) is the distance between the slits. Since \( D \) and \( d \) remain constant in both cases, we can express the relationship between the two scenarios. ### Step 3: Set up the equation for both cases Let: - \( n_1 \) = number of fringes with wavelength \( \lambda_1 = 4000 \, \text{Å} \) (given as 60) - \( n_2 \) = number of fringes with wavelength \( \lambda_2 = 6000 \, \text{Å} \) From the relationship established earlier: \[ n_1 \beta_1 = n_2 \beta_2 \] Substituting the expression for fringe width: \[ n_1 \left( \frac{\lambda_1 D}{d} \right) = n_2 \left( \frac{\lambda_2 D}{d} \right) \] ### Step 4: Simplify the equation Since \( D \) and \( d \) are constant, they cancel out: \[ n_1 \lambda_1 = n_2 \lambda_2 \] ### Step 5: Substitute known values Now we can substitute the known values into the equation: \[ 60 \times 4000 \, \text{Å} = n_2 \times 6000 \, \text{Å} \] ### Step 6: Solve for \( n_2 \) Rearranging the equation gives: \[ n_2 = \frac{60 \times 4000}{6000} \] Calculating this: \[ n_2 = \frac{240000}{6000} = 40 \] ### Conclusion The number of fringes obtained with a wavelength of \( 6000 \, \text{Å} \) in the same field of view is \( 40 \).

To solve the problem, we will follow these steps: ### Step 1: Understand the relationship between the number of fringes and fringe width In Young's double slit experiment, the number of fringes (n) in a given field of view is given by the formula: \[ n = \frac{L}{\beta} \] where \( L \) is the length of the region and \( \beta \) is the fringe width. ### Step 2: Express the fringe width in terms of wavelength ...
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RESONANCE ENGLISH-WAVE OPTICS-Exercise
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  16. Two coherent monochromatic light beams of intensities I and 4I are sup...

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