A light of wavelength 5890 mathringA falls normally on a thin air film. The minimum thickness of the film such that the film appears dark in reflected light is

Question

A light of wavelength 5890 mathringA falls normally on a thin air film. The minimum thickness of the film such that the film appears dark in reflected light is (A) 2.945 × 10-7 m (B) 3.945 × 10-7 m (C) 4.95 × 10-7 m (D) 1.945 × 10-7 m

Tardigrade Admin · Accepted Answer

If thin film appears dark 2 μ t cos r=n λ for normal incidence r=0° ⇒ 2 μ t=n λ ⇒ t=(n λ/2 μ) ⇒ t min =(λ/2 μ)=(5890 × 10-10/2 × 1) =2.945 × 10-7 m

Q. A light of wavelength $5890 \overset{˚}{A}$ falls normally on a thin air film. The minimum thickness of the film such that the film appears dark in reflected light is

$2.945 \times 1 0^{- 7} m$

$3.945 \times 1 0^{- 7} m$

$4.95 \times 1 0^{- 7} m$

$1.945 \times 1 0^{- 7} m$

If thin film appears dark
$2 μ t cos r = nλ$ for normal incidence $r = 0^{\circ}$
$\Rightarrow 2 μ t = nλ \Rightarrow t = \frac{nλ}{2 μ}$
$\Rightarrow t_{m i n} = \frac{λ}{2 μ} = \frac{5890 \times 1 0 ^{- 10}}{2 \times 1}$
$= 2.945 \times 1 0^{- 7} m$

Q. A light of wavelength 5890A˚ falls normally on a thin air film. The minimum thickness of the film such that the film appears dark in reflected light is

2.945×10−7m

3.945×10−7m

4.95×10−7m

1.945×10−7m