Question

Two slits in Young’s experiment have widths in the ratio\[1:25\]. The ratio of intensity at the maxima and the minima in the interference pattern,$\dfrac{I_{max}}{I_{min}}$ is
A.$\dfrac{4}{9}$
B.$\dfrac{9}{4}$
C.$\dfrac{121}{49}$
D.$\dfrac{49}{121}$

Answer 1

Hint: To find the ratio of intensity of max to min interference, we must find the amplitude. To find the amplitude we need to find the intensity which depends on the slit width. We can use the following equations, $I\propto d$,$I\propto a^{2}$ or $\sqrt I\propto a$,\[\dfrac{{{I}_{\max }}}{{{I}_{\min }}}=\dfrac{{{({{a}_{1}}+{{a}_{2}})}^{2}}}{{{({{a}_{1}}-{{a}_{2}})}^{2}}}\]
Formula used:
$I\propto d$
$I\propto a^{2}$ or $\sqrt I\propto a$
\[\dfrac{{{I}_{\max }}}{{{I}_{\min }}}=\dfrac{{{({{a}_{1}}+{{a}_{2}})}^{2}}}{{{({{a}_{1}}-{{a}_{2}})}^{2}}}\]

Complete step-by-step answer:
Young’s interference or Young’s double-slit experiment gave the way to the acceptance of the wave theory of light, in the early 19th century.
An interference pattern is observed when two coherent sources of light are superimposed.The intensity of the superimposed waves is the energy transferred per unit perpendicular area.
Given that the ratio of widths, $\dfrac{d_{1}}{d_{2}}=\dfrac{1}{25}$
We know that the intensity of the interference is proportional to the width of the slits.$I\propto d$
Then, \[\dfrac{{{I}_{_{1}}}}{{{I}_{2}}}=\dfrac{{{d}_{1}}}{{{d}_{2}}}=\dfrac{1}{25}\]
We also know that the intensity is proportional to the square of the amplitude.
$I\propto a^{2}$ or $\sqrt I\propto a$
Then,\[\dfrac{{{I}_{_{1}}}}{{{I}_{2}}}=\dfrac{a_{1}^{2}}{a_{2}^{2}}\]
\[\dfrac{{{a}_{1}}}{{{a}_{2}}}=\sqrt{\dfrac{{{I}_{1}}}{{{I}_{2}}}}=\sqrt{\dfrac{1}{25}}=\dfrac{1}{5}\]
The maximum intensity is given by $I_{max} \propto (a_{1}+a_{2})^{2}$
And the minimum intensity is given by $I_{min} \propto (a_{1}-a_{2})^{2}$
Then the ratio of maximum intensity to minimum intensity is given by
\[\dfrac{{{I}_{\max }}}{{{I}_{\min }}}=\dfrac{{{({{a}_{1}}+{{a}_{2}})}^{2}}}{{{({{a}_{1}}-{{a}_{2}})}^{2}}}=\dfrac{{{(1+5)}^{2}}}{{{(1-5)}^{2}}}=\dfrac{{{(6)}^{2}}}{{{(4)}^{2}}}=\dfrac{36}{16}=\dfrac{9}{4}\]
Hence the answer is B.$\dfrac{9}{4}$

Note:
Remember the formulas used. They are often useful. Note only when two coherent sources of light are superimposed, $I\propto d$ and$I\propto a^{2}$ which means that$a^{2} \propto d$. Also, the maximum intensity is given by $I_{max} \propto (a_{1}+a_{2})^{2}$, while the minimum intensity is given by $I_{min} \propto (a_{1}-a_{2})^{2}$