COMEDK · Physics · 26. Wave Optics
In the young's double slit experiment the fringe width of the interference pattern is found to be \(3.2 \times 10^{-4} \mathrm{~m}\), when the light of wave length \(6400^{\circ} \mathrm{A}\) is used. What will be change in fringe width if the light is replaced with a light of wave length \(4800^{\circ} \mathrm{A}\)
- A \(1.6 \times 10^{-4} \mathrm{~m}\)
- B \(2.4 \times 10^{-4} \mathrm{~m}\)
- C \(5.6 \times 10^{-4} \mathrm{~m}\)
- D \(0.8 \times 10^{-4} \mathrm{~m}\)
Answer & Solution
Correct Answer
(D) \(0.8 \times 10^{-4} \mathrm{~m}\)
Step-by-step Solution
Detailed explanation
The fringe width \(\beta\) in Young's double slit experiment is given by the formula \(\beta = \dfrac{\lambda D}{d}\), where \(\lambda\) is the wavelength of light, \(D\) is the distance between the screen and the slits, and \(d\) is the distance between the two slits. Given the…
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