Question 355: Waves Optics

Question

Two monochromatic beams of intensities I and 4 I respectively are superposed to form a steady interference pattern. The maximum and minimum intensities in the pattern are

Accepted Answer

$$\frac{\mathrm{I}_1}{\mathrm{I}_2}=\frac{\mathrm{I}}{4 \mathrm{I}}=\frac{1}{4}=\frac{\mathrm{a}_1^2}{\mathrm{a}_2^2}$$
where $$a_1$$ and $$a_2$$ are the amplitudes
$$\begin{aligned}
& 	herefore \frac{a_1}{a_2}=\frac{1}{2} \quad 	herefore \frac{a_{\max }}{a_{\min }}=\frac{a_1+a_2}{a_1-a_2}=\frac{1.2}{1-2}=\frac{3}{-1} \
& 	herefore \frac{I_{\max }}{I_{\min }}=\frac{a_{\max }^2}{a_{\min }^2}=\frac{9}{1}
\end{aligned}$$

Answer

$$4 \mathrm{I}$$ and I

Answer

9 I and 3I

Answer

5 I and 3 I

Answer

9 I and I

Waves Optics

Two monochromatic beams of intensities I and 4 I respectively are superposed to form a steady interference pattern. The maximum and minimum intensities in the pattern are

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