Q150 / 150 · MHT CET 2026 (16 April Shift 1) · Physics
In Young's double slit experiment, the fringe width is 0.4 mm. What is the distance between $4^{\text{th}}$ dark band and $6^{\text{th}}$ bright band on the same side of the interference pattern?
- A$0.5$ mm
- B$0.75$ mm
- Correct answerC$1.0$ mm
- D$1.5$ mm
Solution
The fringe width is given as $\beta = 0.4$ mm.
The position of the $n^{\text{th}}$ bright fringe from the central maximum is given by $y_n = n\beta$.
For the $6^{\text{th}}$ bright fringe, $y_6 = 6\beta$.
The position of the $n^{\text{th}}$ dark fringe from the central maximum is given by $y'_n = (n - 0.5)\beta$.
For the $4^{\text{th}}$ dark fringe, $y'_4 = (4 - 0.5)\beta = 3.5\beta$.
The distance between the $4^{\text{th}}$ dark band and the $6^{\text{th}}$ bright band on the same side is:
$\Delta y = y_6 - y'_4$
$\Delta y = 6\beta - 3.5\beta = 2.5\beta$
Substituting the value of $\beta$:
$\Delta y = 2.5 \times 0.4$ mm $= 1.0$ mm
Answer: $1.0$ mm
The position of the $n^{\text{th}}$ bright fringe from the central maximum is given by $y_n = n\beta$.
For the $6^{\text{th}}$ bright fringe, $y_6 = 6\beta$.
The position of the $n^{\text{th}}$ dark fringe from the central maximum is given by $y'_n = (n - 0.5)\beta$.
For the $4^{\text{th}}$ dark fringe, $y'_4 = (4 - 0.5)\beta = 3.5\beta$.
The distance between the $4^{\text{th}}$ dark band and the $6^{\text{th}}$ bright band on the same side is:
$\Delta y = y_6 - y'_4$
$\Delta y = 6\beta - 3.5\beta = 2.5\beta$
Substituting the value of $\beta$:
$\Delta y = 2.5 \times 0.4$ mm $= 1.0$ mm
Answer: $1.0$ mm
Hence, the correct answer is option: 'C'.