When a plastic thin film of refractive index 1.45 is placed in the path of one of the interfering waves then the central fringe is displaced through width of five fringes. The thickness of the film, if the wavelength of light is 5890 mathringA, will be

Question

When a plastic thin film of refractive index 1.45 is placed in the path of one of the interfering waves then the central fringe is displaced through width of five fringes. The thickness of the film, if the wavelength of light is 5890 mathringA, will be (A) 6.544 × 10-4 cm  (B) 6.544 × 10-4 m  (C) 6.54 × 10-4 cm  (D) 6.5 × 10-4 cm

Tardigrade Admin · Accepted Answer

X 0=(β/λ)(μ-1) t ⇒ 5 β=(β(0.45) t /5890 × 10-10) t =(5 × 5890 × 10-10/0.45) =6.544 × 10-4 cm

Q. When a plastic thin film of refractive index $1.45$ is placed in the path of one of the interfering waves then the central fringe is displaced through width of five fringes. The thickness of the film, if the wavelength of light is $5890 \overset{˚}{A}$ , will be

$6.544 \times 1 0^{- 4} c m$

$6.544 \times 1 0^{- 4} m$

$6.54 \times 1 0^{- 4} c m$

$6.5 \times 1 0^{- 4} c m$

$X_{0} = \frac{β}{λ} (μ - 1) t$
$\Rightarrow 5 β = \frac{β ( 0.45 ) t}{5890 \times 1 0 ^{- 10}}$
$t = \frac{5 \times 5890 \times 1 0 ^{- 10}}{0.45}$
$= 6.544 \times 1 0^{- 4} c m$

Q. When a plastic thin film of refractive index 1.45 is placed in the path of one of the interfering waves then the central fringe is displaced through width of five fringes. The thickness of the film, if the wavelength of light is 5890A˚, will be

6.544×10−4cm

6.544×10−4m

6.54×10−4cm

6.5×10−4cm