COMEDK · Physics · 26. Wave Optics
In Young's double slit experiment, the ratio of intensities of light from one slit to the other is \(9: 1\). If Im is the maximum intensity, what is the resultant intensity when they interfere at phase difference \(\phi\) ?
- A \(\dfrac{\operatorname{Im}}{9}\left[1+8 \cos ^2\left(\dfrac{\phi}{2}\right)\right]\)
- B \(\dfrac{\operatorname{Im}}{4}\left[1+3 \cos ^2\left(\dfrac{\phi}{2}\right)\right]\)
- C \(\dfrac{\operatorname{Im}}{4}\left[1+8 \cos ^2\left(\dfrac{\phi}{2}\right)\right]\)
- D \(\dfrac{\operatorname{Im}}{2}\left[4+12 \cos ^2\left(\dfrac{\phi}{2}\right)\right]\)
Answer & Solution
Correct Answer
(B) \(\dfrac{\operatorname{Im}}{4}\left[1+3 \cos ^2\left(\dfrac{\phi}{2}\right)\right]\)
Step-by-step Solution
Detailed explanation
Let the intensities of the two slits be \(I_1\) and \(I_2\). Given the ratio \(I_1 : I_2 = 9 : 1\), we can write \(I_1 = 9k\) and \(I_2 = k\) for some constant \(k\). The maximum intensity \(I_m\) is given by…
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