In Young's double slit experiment using monochromatic light of wavelength 'λ' the intensity of light at a point on the screen is 20 units, where the path difference is λ. The intensity of light at a point having the path difference λ/4 is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 10 units
Given:
Path difference $\delta = \lambda$ → Intensity $I = 20$ units
In Young's double slit experiment, intensity at a point is:
$I = 4 I_0 \cos^2 \frac{\delta}{2}$
At $\delta = \lambda$, $I = 4 I_0 \cos^2 \frac{\lambda}{2}$
Since $\cos^2 \frac{\lambda}{2} = \cos^2 (\pi) = 1$, so $I = 4 I_0 = 20 \;\Rightarrow\; I_0 = 5$
At $\delta = \frac{\lambda}{4}$:
$I = 4 I_0 \cos^2 \frac{\lambda/4}{2} = 4 I_0 \cos^2 \frac{\lambda}{8} = 4 \cdot 5 \cdot \cos^2 \frac{\pi}{4}$
$\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$, so $\cos^2 \frac{\pi}{4} = \frac{1}{2}$
$I = 4 \cdot 5 \cdot \frac{1}{2} = 10$ units
Answer: Intensity $= 10$ units