Q132 / 155 · WBJEE 2023 · Physics
An interference pattern is obtained with two coherent sources of intensity ratio \(\mathrm{n}: 1\). The ratio \(\frac{\mathrm{I}_{\text {Max }}-\mathrm{I}_{\text {Min }}}{\mathrm{I}_{\text {Max }}+\mathrm{I}_{\text {Min }}}\) will be maximum if
- Correct answerA\(n=1\)
- B\(n=2\)
- C\(n=3\)
- D\(n=4\)
Solution
Hint: \(\frac{I_1}{I_2}=n\)
\(\begin{aligned}
& \mathrm{I}_1=\mathrm{nI}_2 \\
& \mathrm{I}_{\max }=\left(\sqrt{\mathrm{I}_1}+\sqrt{\mathrm{I}_2}\right)^2=\left(\sqrt{\mathrm{nI}} 2+\sqrt{\mathrm{I}_2}\right)^2=(\sqrt{\mathrm{n}}+1)^2\left(\sqrt{\mathrm{I}_2}\right)^2
\end{aligned}\)
\(\begin{aligned}
& I_{\min }=\left(\sqrt{I_1}-\sqrt{I_2}\right)^2=\left(\sqrt{n I_2}-\sqrt{I_2}\right)^2=(\sqrt{n}-1)^2\left(\sqrt{I_2}\right)^2 \\
& \frac{I_{\max }-I_{\min }}{I_{\max }+I_{\min }}=\frac{(\sqrt{n}+1)^2\left(I_2\right)-(\sqrt{n}-1)^2 I_2}{(\sqrt{n}+1)^2 I_2+(\sqrt{n}-1)^2 I_2}=\frac{(\sqrt{n}+1)^2-(\sqrt{n}-1)^2}{(\sqrt{n}+1)^2+(\sqrt{n}-1)^2} \\
& =\frac{(n+1+2 \sqrt{n})-(n+1-2 \sqrt{n})}{n+1+2 \sqrt{n}+n+1-2 \sqrt{n}}=\frac{4 \sqrt{n}}{2(n+1)}=\frac{2 \sqrt{n}}{n+1}
\end{aligned}\)
\(\therefore\) decreases with increasing \(\mathrm{n}\).
\(\therefore\) It will be maximum if \(n=1\)
Hence, the correct answer is option: 'A'.