Question

In a Young's double slit arrangement, frings are produced using light of wavelength 4000 Å. One slit is covered by a thin plate of glass of refractive index 1.4 and the other with another glass plate of the same thickness but of refractive index 1.7. By doing so, the central bright fringe shifts to the original sixth fringe from the center. The thickness of the glass plates is:

• A. 2 \( \mu m \)
• B. 8 \( \mu m \)
• C. 11 \( \mu m \)
• D. 16 \( \mu m \)

Hint:

In Young's double-slit experiment, the introduction of different thicknesses and refractive indices of glass plates leads to a phase difference, shifting the central maximum. The path difference \( \Delta {OPD} = t(n-1) \) plays a key role in this shift.

Answer 1

The Correct Option is B

In Young's double-slit experiment, the shift in the central fringe occurs due to the introduction of a phase difference between the two slits. When a glass plate of thickness \( t \) and refractive index \( n \) is introduced, the optical path difference changes. The phase shift introduced by the glass plate is given by: \[ \Delta {OPD} = t \left( n - 1 \right) \] where \( t \) is the thickness of the glass plate and \( n \) is the refractive index of the glass. Step 1: For the first slit with refractive index \( n_1 = 1.4 \) and thickness \( t \), the path difference is: \[ \Delta {OPD}_1 = t \times (1.4 - 1) = 0.4t \] For the second slit with refractive index \( n_2 = 1.7 \) and the same thickness \( t \), the path difference is: \[ \Delta {OPD}_2 = t \times (1.7 - 1) = 0.7t \] The total optical path difference between the two slits is: \[ \Delta {OPD} = \Delta {OPD}_2 - \Delta {OPD}_1 = 0.7t - 0.4t = 0.3t \] 
Step 2: This optical path difference shifts the central maximum to the sixth fringe. The fringe width \( \beta \) is given by: \[ \beta = \frac{\lambda D}{d} \] where \( \lambda \) is the wavelength of light, \( D \) is the distance between the screen and the slits, and \( d \) is the distance between the slits. The shift in the fringe due to the optical path difference is given by: \[ \Delta y = \frac{\Delta {OPD}}{\lambda} \times \beta \] Given that the shift is to the sixth fringe, we can equate the shift to \( 6\beta \), and solve for \( t \). 
Step 3: Using \( \Delta {OPD} = 0.3t \) and substituting into the above equation: \[ 0.3t = 6 \times \lambda \] Substitute \( \lambda = 4000 \, {Å} = 4000 \times 10^{-10} \, {m} = 4 \times 10^{-7} \, {m} \) into the equation: \[ 0.3t = 6 \times 4 \times 10^{-7} \] \[ 0.3t = 2.4 \times 10^{-6} \] \[ t = \frac{2.4 \times 10^{-6}}{0.3} = 8 \times 10^{-6} \, {m} = 8 \, \mu m \] Thus, the thickness of the glass plates is \( \boxed{8 \, \mu m} \).