Subject: Applied Physics 2

Topic: Interference And Diffraction Of Light

Difficulty: Medium

Consider a point A very near to edge of the wedge such that thickness of the film at A is $t_1$. If $t_1$ is of such thickness that 2μ$t_1$cos(r+ѳ) is odd multiple of λ/2 then A will look bright. All along the straight line passing through AA’ has same thickness. therefore, a bright band is obtained at A.Let ꞵ be the distance between two bright bands Consider a point C such that thickness of the film at C is $t_3$.If $t_3$ is of such thickness that 2μ$t_3$cos(r+ѳ) is odd multiple of λ/2 then C will look bright. All along the straight line passing through CC’ has same thickness. therefore, a bright band is obtained at C.Let ꞵ be the distance between two bright bands.

For normal incidence and very small angle cos(r+ѳ) =1

2 μ $t_1$ cos(r+ѳ) = λ/2……………………………………………….(1)

2 μ $t_3$ cos(r+ѳ) =3λ/2………………………………………………..(2)

$t_1$ = λ/4μ $t_3$ = 3λ/4μ

$t_3$ - $t_1$ = λ/2μ

tan ѳ = $t_3$ - $t_1$ /ꞵ

ꞵ= λ/2μ tan ѳ

For very small angle tan ѳ = ѳ

Therefore ꞵ= λ/2μѳ………………………………….(3)

Similarly, for dark bands 2 μ $t_2$ = 2λ/2 i.e $t_2$ = λ/4μ

2 μ $t_4$ = 4λ/2 i.e $t_4$ = λ/μ

tan ѳ = ($t_4$ - $t_2$) /ꞵ

ꞵ= λ/(2μ tan ѳ)

For very small angle tan ѳ = ѳ

Therefore ꞵ= λ/2μѳ………………………………….(4)

From equation 3 and 4, distance between two dark bands and bright bands are same. So bands are equally spaced.

DIAMETER OF A THIN WIRE:

Take two optically flat glass plates and held them together at one end so that a wedge shaped air film is formed with a very small wedge angle ѳ. At other end put the wire or foil whose diameter is to be determined. From the figure, tan ѳ = d/l…………………….(1)

also tan ѳ =λ/2μꞵ…………….(2)

from 1 &2

d/l = λ / 2μꞵ i.e d = lλ / 2ꞵ

for an air film μ =1, ꞵ which is distance between two dark or bright/dark bands can be measured experimentally. Therefore, diameter can be determined.