Q134 / 150 · MHT CET 2023 (09 May Shift 2) · Physics
In Young's double slit experiment, the fringe width is $2 \mathrm{~mm}$. The separation between the $13^{\text {th }}$ bright fringe and the $4^{\text {th }}$ dark fringe from the centre of the screen on same side will be
- A$13 \mathrm{~mm}$.
- B$17 \mathrm{~mm}$.
- Correct answerC$19 \mathrm{~mm}$.
- D$23 \mathrm{~mm}$.
Solution
Given: Fringe width $\mathrm{W}=2 \mathrm{~mm}$ The distance of the $\mathrm{n}^{\text {th }}$ bright fringe from centre of the screen $\mathrm{y}_{\mathrm{n}}=\frac{\mathrm{n} \lambda \mathrm{D}}{\mathrm{d}}$ ... (i)
The distance of the $\mathrm{n}^{\text {th }}$ dark fringe from centre of the screen $y_n^{\prime}=(2 n-1) \frac{\lambda D}{2 d}$ ... (ii)
Substituting, $\mathrm{n}=13$ in (i) and $\mathrm{n}=4$ in (ii) we get $\mathrm{y}_{13}=\frac{13 \lambda \mathrm{D}}{\mathrm{d}}$ and $\mathrm{y}_4^{\prime}=\frac{7}{2} \frac{\lambda \mathrm{D}}{\mathrm{d}}$
The separation between the $13^{\text {th }}$ bright fringe and the $7^{\text {th }}$ dark fringe is
$\begin{aligned} & \quad \mathrm{y}_{13}-\mathrm{y}_4^{\prime} \\ & =\frac{13 \lambda \mathrm{D}}{\mathrm{d}}-\frac{7}{2} \frac{\lambda \mathrm{D}}{\mathrm{d}} \\ & =\left(13-\frac{7}{2}\right) \frac{\lambda \mathrm{D}}{\mathrm{d}} \\ & =\frac{19}{2} \frac{\lambda \mathrm{D}}{\mathrm{d}}=\frac{19}{2} \mathrm{~W} \\ & \\ & \text { but } \mathrm{W}=2 \mathrm{~mm} \\ & \therefore \quad \mathrm{y}_{13}-\mathrm{y}_4^{\prime}=\frac{19}{2} \times 2=19 \mathrm{~mm}\end{aligned}$
Hence, the correct answer is option: 'C'.