In Young's double slit experiment, using monochromatic light of wavelength $\lambda$, the intensity of light at a point on the screen where the path difference is $\lambda$, is $K$ units. The intensity of light at a point where the path difference is $\frac{\lambda}{3}$ will be

Question

Accepted Answer

A. 1. Recall that when two coherent waves of equal intensity $I_0$ superpose, the resultant intensity depends on the phase difference $\Delta\phi$ between them.
$$
I = I_0\cos^2\left(\frac{\Delta\phi}{2}ight)
$$

2. The phase difference $\Delta\phi$ is related to the path difference $\Delta x$ by the standard relation.
$$
\Delta\phi = \frac{2\pi}{\lambda}\Delta x
$$

3. Given that a path difference of $\lambda$ produces intensity $K$, substitute $\Delta x = \lambda$ into the intensity formula.
$$
K = I_0\cos^2\left(\frac{2\pi}{\lambda}\cdot\frac{\lambda}{2}ight) = I_0\cos^2(\pi) = I_0
$$
Therefore $I_0 = K$.

4. Now find the intensity for a path difference of $\frac{\lambda}{3}$ using the same formula.
$$
I_1 = I_0\cos^2\left(\frac{2\pi}{\lambda}\cdot\frac{\lambda}{6}ight) = I_0\cos^2\left(\frac{\pi}{3}ight)
$$

5. Since $\cos\left(\frac{\pi}{3}ight) = \frac12$, simplify the expression.
$$
I_1 = I_0	imes\frac14 = \frac{K}{4}
$$

Hence, the answer is **A**.

In Young's double slit experiment, using monochromatic light of wavelength λ\lambdaλ, the intensity of light at a point on the screen where the path difference is λ\lambdaλ, is KKK units. The intensity of light at a point where the path difference is λ3\frac{\lambda}{3}3λ​ will be

In Young's double slit experiment, using monochromatic light of wavelength $\lambda$ , the intensity of light at a point on the screen where the path difference is $\lambda$ , is $K$ units. The intensity of light at a point where the path difference is $\frac{\lambda}{3}$ will be