TG EAMCET 3 May 2025 Shift 1 Paper

Step 1: Find the position equations for the fringes.The position of the $n$th bright fringe is given by: $y_n = \; \frac{n D \lambda}{d}$So, for the 5th bright fringe: $y_5 = \; \frac{5 D \lambda}{d}$The position of the $n$th dark fringe is: $y_n = \; \frac{(2n-1) D \lambda}{2d}$For the 7 th dark fringe: $y_7 = \; \frac{(2 \times 7 - 1) D \lambda}{2d} = \; \frac{13 D \lambda}{2d}$Step 2: Use the given distance between 5th bright and 7th dark fringes.Distance between the 7th dark and 5th bright fringe: $y_7 - y_5 = \; \frac{13 D \lambda}{2d} - \; \frac{5 D \lambda}{d}$First, make the denominators the same: $\; \frac{13 D \lambda}{2d} - \; \frac{10 D \lambda}{2d} = \; \frac{3 D \lambda}{2d}$The distance given in the question is $3 \text{ mm}: \; \frac{3 D \lambda}{2d} = 3 \times 10^{-3} \text{ m}$Step 3: Find the value of $\; \frac{D \lambda}{d}$.Divide both sides by 3: $\; \frac{D \lambda}{2d} = 1 \times 10^{-3}$ Now multiply both sides by 2 : $\; \frac{D \lambda}{d} = 2 \times 10^{-3} \text{ m}$ So, $\; \frac{D \lambda}{d} = 2 \text{ mm}$Step 4: Find the positions needed for the second distance.Position of the 7th bright fringe: $y_7' = \; \frac{7 D \lambda}{d}$Position of the 5th dark fringe: $y_5' = \; \frac{(2 \times 5 - 1) D \lambda}{2d} = \; \frac{9 D \lambda}{2d}$Step 5: Calculate the distance between 5th dark and 7th bright fringes.$y_7' - y_5' = \; \frac{7 D \lambda}{d} - \; \frac{9 D \lambda}{2d}$ Make denominators the same: $\; \frac{14 D \lambda}{2d} - \; \frac{9 D \lambda}{2d} = \; \frac{5 D \lambda}{2d}$Step 6: Find the final answer using the value from earlier.From (i), $\; \frac{D \lambda}{d} = 2 \text{ mm}$, so: $\; \frac{5 D \lambda}{2d} = \; \frac{5}{2} \times 2 \text{ mm} = 5 \text{ mm}$Final Answer: Distance between 5th dark and 7th bright fringes is $5 \text{ mm}$.