Q58 / 75 · JEE Main 2026 (04 April Shift 1) · Physics
A slit of width $a$ is illuminated by light of wavelength $\lambda$. The linear separation between $1^{st}$ and $3^{rd}$ minima in the diffraction pattern produced on a screen placed at a distance $D$ from the slit system is _____.
- A$\dfrac{D\lambda}{a}$
- B$1.5\dfrac{D\lambda}{a}$
- Correct answerC$2\dfrac{D\lambda}{a}$
- D$3\dfrac{D\lambda}{a}$
Solution
The position of the $n^{th}$ minimum in a single slit diffraction pattern on a screen at distance $D$ is given by $y_n = \dfrac{n D \lambda}{a}$.
For the $1^{st}$ minimum, $n = 1$, so $y_1 = \dfrac{D \lambda}{a}$.
For the $3^{rd}$ minimum, $n = 3$, so $y_3 = \dfrac{3 D \lambda}{a}$.
The linear separation between the $1^{st}$ and $3^{rd}$ minima on the same side of the central maximum is $\Delta y = y_3 - y_1$.
$\Delta y = \dfrac{3 D \lambda}{a} - \dfrac{D \lambda}{a} = 2\dfrac{D \lambda}{a}$.
Answer: $2\dfrac{D\lambda}{a}$
For the $1^{st}$ minimum, $n = 1$, so $y_1 = \dfrac{D \lambda}{a}$.
For the $3^{rd}$ minimum, $n = 3$, so $y_3 = \dfrac{3 D \lambda}{a}$.
The linear separation between the $1^{st}$ and $3^{rd}$ minima on the same side of the central maximum is $\Delta y = y_3 - y_1$.
$\Delta y = \dfrac{3 D \lambda}{a} - \dfrac{D \lambda}{a} = 2\dfrac{D \lambda}{a}$.
Answer: $2\dfrac{D\lambda}{a}$
Hence, the correct answer is option: 'C'.