Optics 6 Question 2
3. In a Young’s double slit experiment, the ratio of the slit’s width is $4: 1$. The ratio of the intensity of maxima to minima, close to the central fringe on the screen, will be
(a) $4: 1$
(b) $25: 9$
(c) $9: 1$
(d) $(\sqrt{3}+1)^{4}: 16$
(2019 Main, 10 April II)
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Answer:
Correct Answer: 3. (c)
Solution:
Key Idea
Amplitude of light is directly proportional to area through which light is passing.
For same length of slits,
$$ \text { amplitude } \propto(\text { width })^{1 / 2} $$
Also,
$$ \text { intensity } \propto(\text { amplitude })^{2} $$
In YDSE, ratio of intensities of maxima and minima is given by
$$ \frac{I _{\max }}{I _{\min }}=\frac{\left(\sqrt{I _1}+\sqrt{I _2}\right)^{2}}{\left(\sqrt{I _1}-\sqrt{I _2}\right)^{2}} $$
where, $I _1$ and $I _2$ are the intensities obtained from two slits, respectively.
$$ \Rightarrow \quad \frac{I _{\max }}{I _{\min }}=\frac{\left(a _1+a _2\right)^{2}}{\left(a _1-a _2\right)^{2}} $$
where, $a _1$ and $a _2$ are light amplitudes from slits 1 and 2 , respectively.
$$ \Rightarrow \quad \frac{I _{\max }}{I _{\min }}=\frac{\left(\sqrt{W _1}+\sqrt{W _2}\right)^{2}}{\left(\sqrt{W _1}-\sqrt{W _2}\right)^{2}} $$
where, $W _1$ and $W _2$ are the widths of slits, respectively.
Here, $\frac{W _1}{W _2}={\frac{a _1}{a _2}}^{2}=\frac{4}{1} \Rightarrow \sqrt{\frac{W _1}{W _2}}=\frac{2}{1}$
So, we have
$$ \begin{aligned} \frac{I _{\max }}{I _{\min }}=\Big(\frac{\sqrt{W _1}+\sqrt{W _2}}{\sqrt{W _1}-\sqrt{W _2}}\Big)^2 & =\Big[\frac{\sqrt{\frac{W _1}{W _2}}+1}{\sqrt{\frac{W _1}{W _2}}-1}\Big] \\ & =\Big(\frac{2+1}{2-1}\Big)^{2}=9: 1 \end{aligned} $$
