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The fringe width in YDSE is 2.4 xx 10^(-...

The fringe width in YDSE is `2.4 xx 10^(-4)m`, when red light of wavelength `6400 Å` is used. By how much will it change, if blue light of wavelength `4000 Å` is used ?

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To solve the problem, we will use the formula for fringe width in Young's Double Slit Experiment (YDSE): \[ \text{Fringe Width} (x) = \frac{\lambda D}{d} \] Where: - \( \lambda \) = wavelength of light - \( D \) = distance between the slits and the screen - \( d \) = distance between the slits Given: - Fringe width with red light (\( x_1 \)) = \( 2.4 \times 10^{-4} \) m - Wavelength of red light (\( \lambda_1 \)) = \( 6400 \) Å = \( 6400 \times 10^{-10} \) m - Wavelength of blue light (\( \lambda_2 \)) = \( 4000 \) Å = \( 4000 \times 10^{-10} \) m ### Step 1: Write the relationship between fringe widths and wavelengths From the formula for fringe width, we can express the fringe widths for both wavelengths: \[ x_1 = \frac{\lambda_1 D}{d} \] \[ x_2 = \frac{\lambda_2 D}{d} \] ### Step 2: Set up the ratio of fringe widths Taking the ratio of the fringe widths: \[ \frac{x_1}{x_2} = \frac{\lambda_1}{\lambda_2} \] ### Step 3: Substitute the known values Substituting the known values into the ratio: \[ \frac{2.4 \times 10^{-4}}{x_2} = \frac{6400 \times 10^{-10}}{4000 \times 10^{-10}} \] ### Step 4: Simplify the right side Calculating the right side: \[ \frac{6400}{4000} = 1.6 \] So we have: \[ \frac{2.4 \times 10^{-4}}{x_2} = 1.6 \] ### Step 5: Solve for \( x_2 \) Now, we can solve for \( x_2 \): \[ x_2 = \frac{2.4 \times 10^{-4}}{1.6} \] Calculating \( x_2 \): \[ x_2 = 1.5 \times 10^{-4} \, \text{m} \] ### Step 6: Calculate the change in fringe width Now, we need to find the change in fringe width: Change in fringe width = \( x_1 - x_2 \) Substituting the values: \[ \text{Change} = 2.4 \times 10^{-4} - 1.5 \times 10^{-4} \] Calculating the change: \[ \text{Change} = 0.9 \times 10^{-4} \, \text{m} \] ### Final Answer The change in fringe width when switching from red light to blue light is: \[ 0.9 \times 10^{-4} \, \text{m} \] ---

To solve the problem, we will use the formula for fringe width in Young's Double Slit Experiment (YDSE): \[ \text{Fringe Width} (x) = \frac{\lambda D}{d} \] Where: - \( \lambda \) = wavelength of light - \( D \) = distance between the slits and the screen - \( d \) = distance between the slits ...
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PRADEEP-OPTICS-Exercise
  1. The ratio of intensities of minima to maxima in Young's double slit ex...

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  2. Yellow light of wavelength 6000Å produces fringes of width 0.8 mm in Y...

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  3. The fringe width in YDSE is 2.4 xx 10^(-4)m, when red light of wavelen...

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  4. State two conditions to obtain sustained interference of light. In you...

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  5. In a Young's expt., the width of the fringes obtained with the light o...

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  6. State two conditions to obtain sustained interference of light. In you...

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  7. The two slits in Young's double slit experiments are separted by a di...

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  8. A double slit is illuminated by light of wave length 6000 Å. The slit ...

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  9. In Young's double-slit experiment the angular width of a fringe formed...

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  10. In Young's double slit experiment, the slits are 0.2 mm apart and the ...

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  11. In a Young's expt., the width of the fringes obtained with the light o...

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  12. In Young's experiment, two coherent sources are 1.5 mm apart and the f...

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  13. A central fringe of interference pattern produced by light of waveleng...

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  14. The interference fringes for sodium light (lambda = 5890 Å) in a doubl...

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  15. Laser light of wavelength 630 nm incident on a pair of slits produces ...

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  16. In YDSE, light of wavelength 5000 Å is used. The third bright band on ...

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  17. In YDSE, the slits are separated by 0.5 mm and screen is placed 1.0 m ...

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  18. In Young's double slit experiment, the two slits 0.15 mm apart are il...

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  19. Two waves of intensity ration 1 : 9 cross eachother at a point. Calcu...

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  20. In Young's experiment, what will be the phase difference and the path ...

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