Question
Question: Find the intensity at a point on a screen in Young’s double slit experiment where the interfering wa...
Find the intensity at a point on a screen in Young’s double slit experiment where the interfering waves of equal intensity have a path difference of (i) 4λ and (ii) 3λ
Solution
Use the value of path difference given to find out the value of phase difference. To figure out the intensity at a point in Young’s double slit experiment, use the formula to find the resultant intensity at a point.
Complete step by step answer:
Given, the intensities of interfering waves are equals that is,
I1=I2=Io (1)
Phase difference can be written as,
ϕ=λ2πΔx
Where λ is the wavelength and Δx is the phase difference.
The resultant intensity at a point is given as
I=I1+I2+2I1I2cosϕ(2)
(i) For path difference Δx=4λ
Phase difference, ϕ = λ2πΔx =λ2π× 4λ = 2π
The resultant intensity at a point is
I=I1+I2+2I1I2cosϕ
Putting the values of I1, I2 and ϕ, we have
I=I0+I0+2I0I0cos \dfrac{\pi }{2} \\\
⇒I= 2I0
(ii) For path difference Δx=3λ
Phase difference, ϕ=λ2πΔx=λ2π×3λ=32π
The resultant intensity at a point is
I=I1+I2+2I1I2cosϕ
Putting the values of I1, I2 and ϕ, we have
I = {I_0} + {I_0} + 2\sqrt {{I_0}{I_0}} \cos \dfrac{{2\pi }}{3} \\\
\Rightarrow I = 2{I_0} + 2{I_0}\left( { - \dfrac{1}{2}} \right) \\\
\Rightarrow I = {I_0} \\\
Therefore, for a path difference of 4λ and 3λ, the resultant intensities at a point are 2I0 and I0 respectively.
Note:
Here, the intensities of the interfering waves are the same but in some cases the intensities might be different. So, before proceeding we should always check whether the interfering waves have same intensities or different. The double-slit experiment is a demonstration that light and matter can display characteristics of both classically defined waves and particles.