MHT CET · Physics · Wave Optics
The intensity of light coming from one of the slits in Young's double slit experiment is double the intensity from the other slit. The ratio of the maximum intensity to the minimum intensity in the interference fringe pattern observed is
- A \(9: 1\)
- B \(34: 1\)
- C \(4: 1\)
- D \(16: 1\)
Answer & Solution
Correct Answer
(B) \(34: 1\)
Step-by-step Solution
Detailed explanation
Two coherent sources of intensities \(I_1\) and \(I_2\) produce,
maximum intensity, \(\mathrm{I}_{\max }=\left(\sqrt{I_1}+\sqrt{I_2}\right)^2\) and minimum intensity, \(I_{\text {min }}=\left(\sqrt{I_1}-\sqrt{I_2}\right)^2\) in an interference pattern.
\(\frac{I_{\text {max }}}{I_{\text {min }}}=\frac{\left(\sqrt{I_1}+\sqrt{I_2}\right)^2}{\left(\sqrt{I_1}-\sqrt{I_2}\right)^2}=\frac{I_1+I_2+2 \sqrt{I_1 I_2}}{I_1+I_2-2 \sqrt{I_1 I_2}}\)
Given \(\mathrm{I}_1=2 \mathrm{I}_2\)
\(\therefore \quad \frac{I_{\max }}{I_{\min }}=\frac{3 \mathrm{I}_2+2 \sqrt{2} \mathrm{I}_2}{3 \mathrm{I}_2-2 \sqrt{2} \mathrm{I}_2}=\frac{3+2 \sqrt{2}}{3-2 \sqrt{2}}=34\)
maximum intensity, \(\mathrm{I}_{\max }=\left(\sqrt{I_1}+\sqrt{I_2}\right)^2\) and minimum intensity, \(I_{\text {min }}=\left(\sqrt{I_1}-\sqrt{I_2}\right)^2\) in an interference pattern.
\(\frac{I_{\text {max }}}{I_{\text {min }}}=\frac{\left(\sqrt{I_1}+\sqrt{I_2}\right)^2}{\left(\sqrt{I_1}-\sqrt{I_2}\right)^2}=\frac{I_1+I_2+2 \sqrt{I_1 I_2}}{I_1+I_2-2 \sqrt{I_1 I_2}}\)
Given \(\mathrm{I}_1=2 \mathrm{I}_2\)
\(\therefore \quad \frac{I_{\max }}{I_{\min }}=\frac{3 \mathrm{I}_2+2 \sqrt{2} \mathrm{I}_2}{3 \mathrm{I}_2-2 \sqrt{2} \mathrm{I}_2}=\frac{3+2 \sqrt{2}}{3-2 \sqrt{2}}=34\)
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