Question 40: Wave Optics

Question

In Young's double slit experiment, the ratio of maximum and minimum intensities in the fringe system is $$9: 1$$. The ratio of amplitudes of coherent sources is

Accepted Answer

Assume, Young's double slit experiment, amplitudes corresponding to the sources $$S_1$$ and $$S_2$$ be $$a_1$$ and $$a_2$$ and intensities be $$I_1$$ and $$I_2$$ respectively.
So, $$I_{\max } \propto a_{	ext {max }}^2$$ and $$I_{	ext {min }} \propto a_{	ext {min }}^2$$
$$\Rightarrow \frac{I_{\max }}{I_{\min }}=\left(\frac{a_{\max }}{a_{\min }}ight)^2=\left(\frac{a_1+a_2}{a_1-a_2}ight)^2$$
According to given question,
$$\begin{aligned}
& \frac{I_{\max }}{I_{\min }}=\frac{9}{1} \
& \Rightarrow \quad 9=\left(\frac{a_1 / a_2+1}{a_1 / a_2-1}ight)^2 \ 
& \Rightarrow \quad(3)^2=\left(\frac{a_1 / a_2+1}{a_1 / a_2-1}ight)^2 \quad 	ext{(given)}\
& \Rightarrow \quad 3=\frac{a_1 / a_2+1}{a_1 / a_2-1} \Rightarrow 3=\frac{\left(\frac{a_1+a_2}{a_2}ight)}{\left(\frac{a_1-a_2}{a_2}ight)} \Rightarrow 3=\frac{a_1+a_2}{a_1-a_2}\
& \Rightarrow \quad 3\left(a_1-a_2ight)=a_1+a_2 \Rightarrow 3 a_1-3 a_2=a_1+a_2 \
&\Rightarrow \quad 2 a_1=4 a_2 \Rightarrow \frac{a_1}{a_2}=\frac{4}{2} \Rightarrow a_1: a_2=2: 1
\end{aligned}$$

Answer

$$9: 1$$

Answer

$$3: 1$$

Answer

$$2: 1$$

Answer

$$1: 1$$

Wave Optics

In Young's double slit experiment, the ratio of maximum and minimum intensities in the fringe system is $9: 1$. The ratio of amplitudes of coherent sources is

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