COMEDK · Physics · 26. Wave Optics
In a YDSE, the distance between the two slits is 2 mm and the wavelength of incident light is 578.9 nm . Path difference between the waves from lower and upper slits reaching a point P where a bright fringe is formed is \(1.36 \mu \mathrm{~m}\). What is the distance of point \(P\) from the central bright fringe if the distance between the screen and the slits is 250 cm ?
- A 1.7 mm
- B 8.5 mm
- C 85 mm
- D 17 mm
Answer & Solution
Correct Answer
(A) 1.7 mm
Step-by-step Solution
Detailed explanation
The path difference \(\Delta x\) at a point \(P\) on the screen in a Young's Double Slit Experiment is given by the formula \(\Delta x = d \sin \theta \approx d \tan \theta = d \dfrac{y}{D}\), where \(d\) is the distance between the slits, \(y\) is the distance of point \(P\)…
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