Question 31: Waves Optics

Question

In Young's double slit experiment, the ratio of intensities of light from one slit to the other is $$9: 1$$. If Im is the maximum intensity, what is the resultant intensity when they interfere at phase difference $$\phi$$ ?

Accepted Answer

In Young's double slit experiment, when light waves from both slits interfere, the resulting intensity is determined by the amplitude and phase difference of the waves. The given ratio of intensities from one slit to the other is $9:1$.
Let $I_1$ and $I_2$ be the intensities from the first and second slit respectively. Since the intensity is proportional to the square of the amplitude, we can write:
$$ I_1 = 9I_2. $$
Let the amplitudes from the two sources be $A_1$ and $A_2$ such that:
$$ A_1^2 = 9A_2^2. $$
This implies:
$$ A_1 = 3A_2. $$
The resultant intensity $I$ when there is a phase difference $\phi$ between the two waves is given by:
$$ I = I_1 + I_2 + 2\sqrt{I_1I_2}\cos\phi. $$
Substitute $I_1 = 9I_2$ into the equation:
$$ I = 9I_2 + I_2 + 2\sqrt{9I_2 \cdot I_2}\cos\phi. $$
This simplifies to:
$$ I = 10I_2 + 6I_2\cos\phi. $$
Maximum intensity $I_m$ occurs when $\phi = 0$, thus:
$$ I_m = (3A_2 + A_2)^2 = 16A_2^2 \ \Rightarrow \ I_m = 16I_2. $$
Express $I$ in terms of $I_m$:
$$ I = I_2(10 + 6\cos\phi) $$
Thus,
$$ I = \frac{I_m}{16}(10 + 6\cos\phi). $$
Rewriting in terms of $\cos^2\left(\frac{\phi}{2}\right)$ using the identity $\cos\phi = 2\cos^2\left(\frac{\phi}{2}\right) - 1$:
$$ I = \frac{I_m}{16}(10 + 6(2\cos^2\left(\frac{\phi}{2}\right) - 1)). $$
Simplify:
$$ I = \frac{I_m}{16}(10 + 12\cos^2\left(\frac{\phi}{2}\right) - 6). $$
$$ I = \frac{I_m}{16}(4 + 12\cos^2\left(\frac{\phi}{2}\right)). $$
Therefore, the correct option is:
$$ \frac{\operatorname{Im}}{4}\left[1+3 \cos ^2\left(\frac{\phi}{2}\right)\right]. $$ which is represented by Option C.

Answer

$$
\frac{\operatorname{Im}}{9}\left[1+8 \cos ^2\left(\frac{\phi}{2}\right)\right]
$$

Answer

$$
\frac{\operatorname{Im}}{4}\left[1+8 \cos ^2\left(\frac{\phi}{2}\right)\right]
$$

Answer

$$
\frac{\operatorname{Im}}{4}\left[1+3 \cos ^2\left(\frac{\phi}{2}\right)\right]
$$

Answer

$$
\frac{\operatorname{Im}}{2}\left[4+12 \cos ^2\left(\frac{\phi}{2}\right)\right]
$$

Waves Optics

In Young's double slit experiment, the ratio of intensities of light from one slit to the other is $9: 1. If Im is the maximum intensity, what is the resultant intensity when they interfere at phase difference \phi$ ?

Choose the correct answer: