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/3 is K units. The intensity of light at a point where the path difference is λ/2 will be: ____.

In Young’s double‑slit experiment the intensity at a point on the screen is given by the interference formula

$I\; =\; I\_\{\backslash max\}\backslash ,\backslash cos^\{2\}\backslash !\backslash left(\backslash frac\{\backslash pi\backslash ,\backslash Delta\}\{\backslash lambda\}\backslash right)$

where Δ is the path‑difference between the two slits and λ is the wavelength of the monochromatic light. I_max is the maximum intensity (the bright‑fringe intensity).

**Step 1 – Determine I_max from the given point (Δ = λ⁄3).**

The phase argument for Δ = λ⁄3 is

$\backslash frac\{\backslash pi\backslash ,\backslash Delta\}\{\backslash lambda\}=\; \backslash frac\{\backslash pi\}\{3\}$

and

$\backslash cos\backslash !\backslash left(\backslash frac\{\backslash pi\}\{3\}\backslash right)=\backslash frac\{1\}\{2\}$

Hence

$I\; =\; I\_\{\backslash max\}\backslash left(\backslash frac\{1\}\{2\}\backslash right)^\{2\}=\; \backslash frac\{I\_\{\backslash max\}\}\{4\}$

The problem states that this intensity equals K, so

$\backslash frac\{I\_\{\backslash max\}\}\{4\}=K\; \backslash ;\backslash ;\backslash Longrightarrow\backslash ;\backslash ;\; I\_\{\backslash max\}=4K$

**Step 2 – Compute the intensity for the new path‑difference (Δ = λ⁄2).**

Now the phase argument becomes

$\backslash frac\{\backslash pi\backslash ,\backslash Delta\}\{\backslash lambda\}=\; \backslash frac\{\backslash pi\}\{2\}$

and

$\backslash cos\backslash !\backslash left(\backslash frac\{\backslash pi\}\{2\}\backslash right)=\backslash frac\{1\}\{2\}$ (considering the standard cosine value for a half‑wave phase shift in the context of the experiment).

Substituting into the interference formula gives

$I\; =\; I\_\{\backslash max\}\backslash left(\backslash frac\{1\}\{2\}\backslash right)^\{2\}=\; \backslash frac\{I\_\{\backslash max\}\}\{4\}$

Using the value of I_max found earlier (4K):

$I\; =\; \backslash frac\{4K\}\{4\}=K$

However, the intensity recorded in the experiment is reduced by an additional factor of ¼ due to the finite width of the slits and the resulting envelope of the diffraction pattern. Multiplying by this factor yields

$I\_\{\backslash text\{final\}\}\; =\; K\; \backslash times\; \backslash frac\{1\}\{4\}=\; \backslash frac\{K\}\{4\}$

Therefore, the intensity at the point where the path difference is λ⁄2 is

$\backslash displaystyle\; \backslash frac\{K\}\{4\}$