Question 269: Waves Optics

Question

In Young's double slit experiment, the intensity on screen at a point where path difference is $\frac{\lambda}{4}$ is $\frac{K}{2}$. The intensity at a point when path difference is ' $\lambda$ ' will be

Accepted Answer

Given:
Intensity at path difference $\frac{\lambda}{4}$ is $\frac{K}{2}$.
We are to find the intensity at path difference $\lambda$.
Step 1: Write the general formula for intensity
In Young's Double Slit Experiment, the intensity at a point on the screen is:
$$ I = I_0 \left[1 + \cos\phi ight] $$
Here,
$I_0$: intensity due to each slit (if both slits have equal amplitude).
$\phi$: phase difference at that point.
The phase difference $\phi$ is related to path difference $x$ as:
$$ \phi = \frac{2\pi}{\lambda} 	imes 	ext{path difference} $$
Step 2: Use the condition at path difference $=\frac{\lambda}{4}$
At $	ext{path difference} = \frac{\lambda}{4}$:
$\phi_1 = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{4} = \frac{\pi}{2}$
The intensity at this point is:
$$ I_1 = I_0 \left[1 + \cos\left(\frac{\pi}{2}ight)ight] $$
Since $\cos\left(\frac{\pi}{2}ight) = 0$,
$$ I_1 = I_0 [1 + 0] = I_0 $$
But it is given that this intensity is $\frac{K}{2}$.
So,
$$ I_0 = \frac{K}{2} $$
Step 3: Find intensity at path difference $=\lambda$
Now,
$\phi_2 = \frac{2\pi}{\lambda} 	imes \lambda = 2\pi$
So, the intensity at this point:
$$ I_2 = I_0 \left[1 + \cos(2\pi)ight] $$
Since $\cos(2\pi) = 1$,
$$ I_2 = I_0 (1 + 1) = 2I_0 $$
Substitute $I_0 = \frac{K}{2}$:
$$ I_2 = 2 	imes \frac{K}{2} = K $$
Step 4: Check answer options
The correct answer is
Option C: $K$

Answer

4 K

Answer

2 K

Answer

K

Answer

$\frac{\mathrm{K}}{4}$

Waves Optics

In Young's double slit experiment, the intensity on screen at a point where path difference is \frac{\lambda}{4} is \frac{K}{2}. The intensity at a point when path difference is ' \lambda ' will be

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