A YDSE is performed with bi-chromatic light (5500 mathringA and 6000 mathringA) for d=2 mm and D=1 m. Distance of first complete maxima from the central maxima on the screen, is

Question

A YDSE is performed with bi-chromatic light (5500 mathringA and 6000 mathringA) for d=2 mm and D=1 m. Distance of first complete maxima from the central maxima on the screen, is (A) 1.1 mm (B) 2.2 mm (C) 3.3 mm (D) 4.4 mm

Tardigrade Admin · Accepted Answer

y=(n1 λ1 D/d)=(n2 λ2 D/d) (n1/n2)=(λ2/λ1)=(12/11) n1=12 y=(12 × 5500 × 10 × 1/2 × 10-3) =330 × 10-5=3.3 mm

Q. A YDSE is performed with bi-chromatic light $(5500 \overset{˚}{A}$ and $6000 \overset{˚}{A})$ for $d = 2 mm$ and $D = 1 m$ . Distance of first complete maxima from the central maxima on the screen, is

1.1 mm

2.2 mm

3.3 mm

4.4 mm

$y = \frac{n _{1} λ _{1} D}{d} = \frac{n _{2} λ _{2} D}{d}$
$\frac{n _{1}}{n _{2}} = \frac{λ _{2}}{λ _{1}} = \frac{12}{11}$
$n_{1} = 12$
$y = \frac{12 \times 5500 \times 10 \times 1}{2 \times 1 0 ^{- 3}}$
$= 330 \times 1 0^{- 5} = 3.3 mm$

Q. A YDSE is performed with bi-chromatic light (5500A˚ and 6000A˚) for d=2mm and D=1m. Distance of first complete maxima from the central maxima on the screen, is