Question

In a Young’s double slit experiment, the path difference at a certain point on the screen between two interfering waves is \(\frac{1}{8}^{th}\) of wavelength. The ratio of the intensity at this point to that at the centre of a bright fringe is close to
1. 0.80
2. 0.74
3. 0.94
4. 0.85

Answer 1

Correct Answer - Option 4 : 0.85

Concept:

Phase Difference:

Phase Difference is used to describe the difference in degrees or radians when two or more alternating quantities reach their maximum or zero values. Phase Difference (ϕ) between two particles or two waves tells us how much a particle (or wave) is in front or behind another particle (or wave).

Phase difference formula is given by

\(\phi = {\rm{\Delta }}x \times \frac{{2\pi }}{\lambda }\) 

Calculation:

Given,

Path difference = 1/8 th of its wavelength.

i.e., \({\rm{\Delta }}x = \frac{\lambda }{8}\)

Let intensity of each wave is I₀

Then, the intensity at the centre of bright fringe will be 4I₀

Given, path difference,

\({\rm{\Delta }}x = \frac{\lambda }{8}\) 

∴ Phase difference,

\(\phi = {\rm{\Delta }}x \times \frac{{2\pi }}{\lambda }\) 

\(\phi = \frac{\lambda }{8} \times \frac{{2\pi }}{\lambda } = \frac{{2\pi }}{8}\) 

\(\phi = \frac{\pi }{4}\) 

Intensity of light at this point,

\(I' = {I_0} + {I_0} + 2{I_0}\cos \left( {\frac{\pi }{4}} \right)\) 

\(I' = 2{I_0} + 2{I_0} \times \frac{1}{{\sqrt 2 }}\) 

\(I' = 2{I_0} + \sqrt 2 {I_0}\) 

\(I' = {I_0}\left( {2 + \sqrt 2 } \right)\) 

I' = I₀ (2 + 1.414) = I₀ (3.414)

I' = 3.414 I₀

Now,

\(\frac{{I'}}{{4{I_0}}} = \frac{{3.41{I_0}}}{{4{I_0}}}\) 

\(\frac{{I'}}{{4{I_0}}} = \frac{{3.414}}{4}\) 

\(\frac{{I'}}{{4{I_0}}} = 0.85\) 

Therefore, the ratio of the intensity at this point to that at the centre of a bright fringe is close to 0.85.