Q179 / 200 · MHT CET - Medical 2026 (23 April Shift 2) · Physics
In a diffraction pattern due to a single slit of width a, the first minimum is observed at an angle $30^\circ$ when light of wavelength $5000$ Å is incident on the slit. The first secondary maximum is observed at an angle of
$\{\sin(30)^\circ = \cos(60)^\circ = \dfrac{1}{2}\}$
$\{\sin(30)^\circ = \cos(60)^\circ = \dfrac{1}{2}\}$
- A$\sin^{-1}\left(\dfrac{2}{3}\right)$
- B$\sin^{-1}\left(\dfrac{1}{2}\right)$
- Correct answerC$\sin^{-1}\left(\dfrac{3}{4}\right)$
- D$\sin^{-1}\left(\dfrac{1}{4}\right)$
Solution
The condition for the first minimum in a single slit diffraction pattern is given by $a \sin \theta = \lambda$.
Substituting $\theta = 30^\circ$, we get $a \sin 30^\circ = \lambda$.
Since $\sin 30^\circ = \dfrac{1}{2}$, we have $a \left(\dfrac{1}{2}\right) = \lambda$, which gives $a = 2\lambda$.
The condition for the first secondary maximum is given by $a \sin \theta' = \dfrac{3\lambda}{2}$.
Substituting $a = 2\lambda$, we get $2\lambda \sin \theta' = \dfrac{3\lambda}{2}$.
Solving for $\sin \theta'$, we get $\sin \theta' = \dfrac{3}{4}$.
Therefore, $\theta' = \sin^{-1}\left(\dfrac{3}{4}\right)$.
Answer: $\sin^{-1}\left(\dfrac{3}{4}\right)$
Substituting $\theta = 30^\circ$, we get $a \sin 30^\circ = \lambda$.
Since $\sin 30^\circ = \dfrac{1}{2}$, we have $a \left(\dfrac{1}{2}\right) = \lambda$, which gives $a = 2\lambda$.
The condition for the first secondary maximum is given by $a \sin \theta' = \dfrac{3\lambda}{2}$.
Substituting $a = 2\lambda$, we get $2\lambda \sin \theta' = \dfrac{3\lambda}{2}$.
Solving for $\sin \theta'$, we get $\sin \theta' = \dfrac{3}{4}$.
Therefore, $\theta' = \sin^{-1}\left(\dfrac{3}{4}\right)$.
Answer: $\sin^{-1}\left(\dfrac{3}{4}\right)$
Hence, the correct answer is option: 'C'.