Question

Two coherent monochromatic light beams of intensities $4I$ and $9I$ interfere in a Young’s double slit experimental setup to produce a fringe pattern on the screen. The phase difference between the beams at two points $P$ and $Q$ on the screen are $\dfrac{\pi }{2}$ and $\dfrac{\pi }{3}$ respectively. Then the ratio of the two intensities $\dfrac{{{I_P}}}{{{I_Q}}}$ is
A.$0$
B. $\dfrac{6}{{19}}$
C. $\dfrac{{13}}{{19}}$
D. $\dfrac{6}{{13}}$

Answer 1

First, we will find the ratio of the amplitudes of the two light beams using the proportionality relation between intensity and amplitude. Using it, the amplitudes at points $P$ and $Q$ is found out. Then, the light intensities at $P$ and $Q$ as well as the ratio will be found from the amplitudes.

Complete step by step answer:
Given: The intensity of the first beam, ${I_1} = 4I$. The intensity of the second beam, ${I_2} = 9I$.The phase difference between the two beams at point $P$, ${\phi _1} = \dfrac{\pi }{2}$.
The phase difference between the two beams at point $Q$, ${\phi _2} = \dfrac{\pi }{3}$.

Let ${A_1}$ and ${A_2}$ be the amplitudes of the first beam and the second beam respectively. As we know, the intensity of a light beam is proportional to the square of its amplitude.Therefore, using the proportionality relation, we can write
$\dfrac{{{I_1}}}{{{I_2}}} = \dfrac{{A_1^2}}{{A_2^2}}$
Substitute the values of ${I_1}$ and ${I_2}$ in the above equation to obtain the ratio of the amplitudes.

$$\dfrac{{4I}}{{9I}} = \dfrac{{A_1^2}}{{A_2^2}}\\\ \Rightarrow\dfrac{{A_1^2}}{{A_2^2}} = \dfrac{4}{9}\\\ \Rightarrow\dfrac{{{A_1}}}{{{A_2}}} = \dfrac{2}{3}$$

It implies ${A_1} = 2a$ and ${A_2} = 3a$, where $a$ is a constant.
The amplitude of light at the point $P$ can be written as,
${A_P} = \sqrt {A_1^2 + A_2^2 + 2{A_1}{A_2}\cos {\phi _1}}$
Now, we substitute the values of ${A_1}$, ${A_2}$ and ${\phi _1}$ in the above equation to get the amplitude at $P$.
${A_P} = \sqrt {{{\left( {2a} \right)}^2} + {{\left( {3a} \right)}^2} + 2 \times 2a \times 3a\cos \dfrac{\pi }{2}} \\\ \Rightarrow{A_P} = \sqrt {4{a^2} + 9{a^2} + 12{a^2} \times 0} \\\ \Rightarrow{A_P} = \sqrt {13} a$
The amplitude of light at the point $Q$ can be written as,
${A_Q} = \sqrt {A_1^2 + A_2^2 + 2{A_1}{A_2}\cos {\phi _2}}$
Now, we substitute the values of ${A_1}$, ${A_2}$ and ${\phi _2}$ in the above equation to get the amplitude at $Q$.
${A_Q} = \sqrt {{{\left( {2a} \right)}^2} + {{\left( {3a} \right)}^2} + 2 \times 2a \times 3a\cos \dfrac{\pi }{3}} \\\ \Rightarrow{A_Q} = \sqrt {4{a^2} + 9{a^2} + 12{a^2} \times \dfrac{1}{2}} \\\ \Rightarrow{A_Q} = \sqrt {19} a$
Therefore, the ratio of the intensities of light at the points $P$ and $Q$ can be written as,
$\dfrac{{{I_P}}}{{{I_Q}}} = \dfrac{{A_P^2}}{{A_Q^2}}$
Here, ${I_P}$ is the intensity at the point $P$ and ${I_Q}$ is the intensity at the point $Q$.
We will put the obtained values of ${A_P}$ and ${A_Q}$ in the above equation to find the intensity ratio.
$\dfrac{{{I_P}}}{{{I_Q}}} = \dfrac{{{{\left( {\sqrt {13} a} \right)}^2}}}{{{{\left( {\sqrt {19} a} \right)}^2}}}\\\ \therefore\dfrac{{{I_P}}}{{{I_Q}}} = \dfrac{{13}}{{19}}$
The ratio of the intensities $\dfrac{{{I_P}}}{{{I_Q}}}$ is obtained as $\dfrac{{13}}{{19}}$.

Hence, option C is the correct answer.

Note: It is not necessary to find the expressions for the amplitudes of the beams to obtain the intensity ratio $\dfrac{{{I_P}}}{{{I_Q}}}$. The intensity of a light beam at any point $P$ can be directly written as ${I_P} = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}} \cos {\phi _1}$. Similarly, the intensity of the beam at the point $Q$ can be expressed as ${I_Q} = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}} \cos {\phi _2}$. We can directly substitute the given values in the intensity equations and then obtain the intensity ratio $\dfrac{{{I_P}}}{{{I_Q}}}$.