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When one of the slits of Young's experim...

When one of the slits of Young's experiment is covered with a transparent sheet of thickness `4.8mm`, the central fringe shifts to a position originally occupied by the 30th bright fringe. What should be the thickness of the sheet if the central fringe has to shift to the position occupied by 20th bright fringe?

A

3.8 mm

B

1.6 mm

C

7.6 mm

D

3.2 mm

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The correct Answer is:
To solve the problem, we need to understand how the thickness of a transparent sheet affects the position of the fringes in Young's double-slit experiment. ### Step-by-Step Solution: 1. **Understanding the Shift of the Central Fringe**: When a transparent sheet of thickness \( t \) and refractive index \( \mu \) is placed in front of one of the slits, it causes a shift in the interference pattern. The optical path difference introduced by the sheet is given by: \[ \Delta = \mu t - t = ( \mu - 1 ) t \] 2. **Relating the Shift to Fringe Order**: The shift in the central fringe corresponds to a certain number of bright fringes. If the central fringe shifts to the position of the 30th bright fringe, we can express this as: \[ \Delta = \frac{30 \lambda}{2} \] where \( \lambda \) is the wavelength of light used. 3. **Setting Up the Equation**: From the above, we can write: \[ (\mu - 1) t_1 = 15 \lambda \] where \( t_1 = 4.8 \, \text{mm} \) is the thickness of the sheet that caused the shift to the 30th fringe. 4. **Finding the New Thickness for the 20th Fringe**: We want to find the thickness \( t_2 \) that will cause the central fringe to shift to the position of the 20th bright fringe. The equation for this shift will be: \[ (\mu - 1) t_2 = 10 \lambda \] 5. **Relating the Two Thicknesses**: Since \( \mu - 1 \) is constant for the same material, we can relate the two equations: \[ \frac{t_2}{t_1} = \frac{10 \lambda}{15 \lambda} = \frac{10}{15} = \frac{2}{3} \] 6. **Calculating \( t_2 \)**: Now substituting \( t_1 = 4.8 \, \text{mm} \) into the equation: \[ t_2 = t_1 \cdot \frac{2}{3} = 4.8 \cdot \frac{2}{3} = 3.2 \, \text{mm} \] ### Final Answer: The thickness of the sheet required for the central fringe to shift to the position occupied by the 20th bright fringe is \( 3.2 \, \text{mm} \). ---
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