Question

In Young's double slit experiment, the ratio of amplitude of light coming from two slits is 2:3. If I₀ be the maximum intensity the resultant intensity I when they interfere at path difference λ/3 (λ = wavelength of light) will be

• A. $\frac{3I_0}{25}$
• B. $\frac{6I_0}{25}$
• C. $\frac{7I_0}{25}$
• D. $\frac{9I_0}{25}$

Answer 1

Answer: (C)

$\frac{7I_0}{25}$

Answer 2

Let the amplitudes of light coming from the two slits be $A_1$ and $A_2$. Given the ratio of amplitudes is $A_1 : A_2 = 2 : 3$. Let $A_1 = 2a$ and $A_2 = 3a$ for some constant $a$.

The intensity of light is proportional to the square of its amplitude ($I \propto A^2$). The maximum intensity ($I_0$) occurs when the waves interfere constructively. The resultant amplitude for constructive interference is $A_{max} = A_1 + A_2$. $A_{max} = 2a + 3a = 5a$. The maximum intensity $I_0$ is proportional to $A_{max}^2$: $I_0 \propto (5a)^2 = 25a^2$. Let $I_0 = K(25a^2)$ for some proportionality constant $K$.

The path difference is given as $\Delta x = \frac{\lambda}{3}$. The phase difference $\phi$ is related to the path difference by the formula: $\phi = \frac{2\pi}{\lambda} \Delta x$ Substituting the given path difference: $\phi = \frac{2\pi}{\lambda} \left(\frac{\lambda}{3}\right) = \frac{2\pi}{3}$.

The resultant amplitude $A$ when two waves with amplitudes $A_1$, $A_2$ and phase difference $\phi$ interfere is given by: $A^2 = A_1^2 + A_2^2 + 2A_1 A_2 \cos \phi$ Substituting the values $A_1=2a$, $A_2=3a$, and $\phi = \frac{2\pi}{3}$: $A^2 = (2a)^2 + (3a)^2 + 2(2a)(3a) \cos\left(\frac{2\pi}{3}\right)$ $A^2 = 4a^2 + 9a^2 + 12a^2 \cos\left(\frac{2\pi}{3}\right)$ Since $\cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}$: $A^2 = 4a^2 + 9a^2 + 12a^2 \left(-\frac{1}{2}\right)$ $A^2 = 13a^2 - 6a^2$ $A^2 = 7a^2$.

The resultant intensity $I$ is proportional to $A^2$: $I \propto 7a^2$. So, $I = K(7a^2)$.

We need to express $I$ in terms of $I_0$. We have $I_0 = K(25a^2)$. From this, we can write $Ka^2 = \frac{I_0}{25}$. Substituting this into the expression for $I$: $I = 7 \left(Ka^2\right) = 7 \left(\frac{I_0}{25}\right) = \frac{7I_0}{25}$.