Question

A Soap Film Of Refractive Index 1.43 Is Illuminated By White Light Incident At An Angle Of 30°. The Refractive Light Is Examined By A Spectroscope In Which Dark Band Corresponding To The Wavelength 600nm Is Observed. Calculate The Thickness Of The Film.

Answer 1

refractive index (u)=1.43

angle of incidence (i)=30

wavelength (w)=600 nm=6×10^-7m

according to snell's law,

u=sin i/sin r

sin r=1.43/1/2

sin r=1.43×2

sin r=2.86

now, as per the formula

2utcosr=w

cos r=√1-(2.86)²=√1-8.17=√-7.17=2.67

t=w/2×u×cos r

t=6×10^-7/2×1.43×2.67

t=6×10^-7/7.63

t=0.78×10^-7m

I hope this helps ( ╹▽╹ )

Answer 2

Answer:

The thickness of soap film is given by $2.24×10^{-7}$ m.

Explanation:

The formula for destructive interference in the thin film in refracted light is given by

2μt cos r = λ (1)

where μ is the refractive index, t is the thickness of film, r is the angle of refraction and λ is the wavelength.

By Snell's law,

sin i/sin r = μ

where i is angle of incidence, r is angle of refraction and μ is refractive index.

Angle of incidence, i = 30°

Refractive index of soap film, μ = 1.43

sin i/sin r = μ

sin r = sin i/μ

sin r = sin 30°/1.43

sin r = 1/(2×1.43) = 0.349

$cos^{2} r = 1 - (0.349)^{2}$

cos r = 0.937

The value of wavelength is given by 600 nm.

From equation (1)

$2μt 0.937 = 600×10^{-9}$

$t = 6×10^{-7}/(0.937×2×1.43)$

$t = 2.24×10^{-7}$m

Therefore, the thickness of soap film = $2.24×10^{-7}$m