Question

How can wavelength of monochromatic light be found from ydse?

Answer 1

Wave Optics In Young's double-slit experiment using monochromatic light of wavelength λ, the intensity of light at a point on the screen where path difference is λ, is K units Monochromatic light of wavelength 589 NM is incident from air on a water surface.

Let I 1= I 2 =I

Let ϕ be the difference between two light waves .

Therefore the resultant intensity

IR=I+I

=2 I + 2 I

=4 I

But 4 I = k

When path difference is λ3 phase difference is 2π3 radius

I′R=I+I+ 2 I 1−−√

I′R=2 I+2 I(−1/2)

=2 I−I

=I

I=k 4

or

Let I 1 and I 2 be the intensity of the two light waves.

For monochromatic light waves, I 1 = I 2

If, Ø = Phase difference between the two waves

Then, their resultant intensities can be obtained by the relation:

Phase difference of the light is given by: 2¶/λ x Path Difference

When path difference = λ,

Phase difference, Ø = 2¶

∴ I' = 2 I 1 +2 I 1 =  4 I 1

Given, I’ = K

∴ I 1 = K/4

When path difference = λ/3

Phase difference, ø = 2¶/3

Hence, resultant intensity is given by,

Using equation (1), we can write:

IR = I 1 = K/4

Hence, the intensity of light with path difference of λ/3 is K/4 units.