Question
The two coherent sources of intensity ratio $\beta$ produce interference. The fringe visibility will be(A) $2 \beta$(B) $(\beta / 2)$(C) $\{\sqrt{\beta} /(1+\beta)\}$(D) $\{(2 \sqrt{\beta}) /(1+\beta)\}$
Step 1
Step 1: The fringe visibility is given by the formula: \[V = \frac{I_{max} - I_{min}}{I_{max} + I_{min}}\] where \(I_{max}\) is the maximum intensity and \(I_{min}\) is the minimum intensity. Show more…
Show all steps
Your feedback will help us improve your experience
Mirza Aslam Beig and 73 other Physics 103 educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Two coherent sources of intensity ratio 1 : 4 produce an interference pattern. The fringe visibility will be (a) 1 (b) 0.8 (c) 0.4 (d) 0.6
The ratio of intensities of rays emitted from two different coherent Sourees is $\alpha .$ For the interference pattern by them, $\left[\left(I_{\max }+I_{\min }\right) /\left(I_{\max }-I_{\min }\right)\right]$ will be equal to (A) $\{(1+\sqrt{\alpha}) / 2 \alpha\}$ (B) $\{(1+\alpha) / 2 \alpha\}$ (C) $\{(1+\sqrt{\alpha}) / 2\}$ (D) $\{(1+\alpha) /(2 \sqrt{\alpha})\}$
Two coherent sources of intensities $I_{1}$ and $I_{2}$ produce an interference pattern. The maximum intensity in the interference pattern will be (a) $I_{1}+I_{2}$ (b) $l_{1}^{2}+l_{2}^{2}$ (c) $\left(l_{1}+l_{2}\right)^{2}$ (d) $\left(\sqrt{I_{1}}+\sqrt{h}\right)^{2}$
Wave Optics
Round 1
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD