In a YDSE, wavelengths 500 nm and 700 nm are used and D/d=10³. The nearest point from the central maximum where their bright fringes coincide again is n×10⁻¹ mm. Find n.
Physics · Wave Optics · young's dubble slit experiments
JEE Mains

Question

In a YDSE, wavelengths 500 nm and 700 nm are used and D/d=10³. The nearest point from the central maximum where their bright fringes coincide again is n×10⁻¹ mm. Find n.

G1 Physics · Accepted Answer

Coincident bright fringes satisfy m₁λ₁=m₂λ₂. The smallest integers are m₁=7 and m₂=5 because 7×500=5×700. Thus y=m₁λ₁D/d=7×500×10⁻⁹×10³=3.5×10⁻³ m=3.5 mm=35×10⁻¹ mm. Therefore n=35.