
θ1=sin−1(a2λ)θ2=sin−1(a3λ)∵θ1+θ2=30∘
⇒sin−1(a2λ)+sin−1(a3λ)=6π⇒a2λ1−(a3λ)2+a3λ1+(a2λ)2=sin6π
Here λ=628 nm
After solving
A=6.07μ m
Approximate Method :
θ=θ1+θ2⇒6π=a2λ+a3λ⇒6π=a5(628 nm)⇒a=6μ m
JEE Main 2025 — Physics Optics
If the measured angular separation between the second minimum to the left of the central maximum and the third minimum to the right of the central maximum is 30∘ in a single slit diffraction pattern recorded using 628 nm light, then the width of the slit is _______ μm.
Held on 2 Apr 2025 · Verified 6 Jul 2026.
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