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Calculate the thickness of soap film of RI 1.43 if a $1^{st}$ order dark band of $\lambda = 6 \times 10^{-7}$ m is observed in the refracted light corresponding to an incidence angle of $30^0$.
written 6.7 years ago by gravatar for neeta.vanage neeta.vanage • 860 modified 2.7 years ago by gravatar for pedsangini276 pedsangini276 • 4.8k

Subject: Applied Physics 2

Topic: Interference And Diffraction

Difficulty: Medium
engineering physics
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written 6.6 years ago by gravatar for neeta.vanage neeta.vanage • 860 modified 6.6 years ago by gravatar for mananbothra mananbothra • 100

μ =1.33, n=1, λ = 6 × 10$^{-7}$ m, thickness t= ?, i= 30

$ μ= \frac{sin \ i}{sin \ r} \\[2ex] sin \ r = \frac{sin \ 30}{1.43} $

sin r = 0.34

Therefore r=20.46

cos r = 0.936

For dark bands 2μ t cos r = nλ

Thickness $t = \frac{nλ}{ 2μ \ cos \ r}$

Therefore $t = \frac{1 \times 6 \times 10^{-7} }{ 2\times1.43 \times0.936}$

Thickness = 2247 AU
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