In Young’s double-slit experiment using monochromatic light of wavelength λ, the intensity of light at a point on the screen where the path difference is λ, is K units. What is the intensity of light at a point where path difference is λ/3?

Hint: Write the way we can represent the intensity of light when path difference equals to lambda, now write the formula for intensity at lambda distance and then replace delta n with the given value of path difference, now find the result and that will be equal to the intensity of light when path difference equals to lambda.

Complete step by step answer:
If path difference is equals to λ wavelength then we represent the intensity of the light as,
${{I}_{\circ }}=K$ ,
We know that the formula for intensity when the intensity at λ distance is ${{I}_{\circ }}$,
Therefore ,
\[I={{I}_{\circ }}{{\cos }^{2}}\dfrac{\Phi }{2}\]…….. eq.1
Now we have to find the phase difference,
We know that the formula for phase difference is,
$\Delta \phi =\dfrac{2\pi }{\lambda }\times \Delta n$ , here n is the path difference,
Here n is given as λ/3 so,
$\Delta \phi =\dfrac{2\pi }{\lambda }\times \dfrac{\lambda }{3}$,
Therefore, $\Delta \phi =\dfrac{2\pi }{3}$,
Now replacing eq.1 with the value of $\Delta \phi$ we get,
\[I={{I}_{\circ }}{{\cos }^{2}}\left( \dfrac{{\scriptstyle{}^{2\pi }/{}_{3}}}{2} \right)\],
\[I={{I}_{\circ }}{{\cos }^{2}}\left( \dfrac{\pi }{3} \right)\],
Now, we know that ${{I}_{\circ }}=K$, and $\cos {{60}^{\circ }}=\dfrac{1}{2}$ ,
$I=K\times {{\left( \dfrac{1}{2} \right)}^{2}}$ ,
I=K/4.
Therefore the intensity of light at a point where path difference is λ/3 is K/4.

Additional Information:
When a light is going through a unit area then intensity I is the power of light going through that area.(in other words, the amount of energy that arrives per unit area, per unit time is known as intensity).

Note:
Try to figure out proper equations from the question that is given, as those equations are the relation from which the correct answer can be reached. In the equation,
\[I={{I}_{\circ }}{{\cos }^{2}}\dfrac{\Phi }{2}\], \[\Phi\] is the phase difference.
We have to find the phase difference from the path difference.