COMEDK · Physics · 26. Wave Optics
In Young's double slit experiment, the two slits are separated by \(0.2 \mathrm{~mm}\) and they are \(1 \mathrm{~m}\) from the screen. The wavelength of the light used is \(500 \mathrm{~nm}\). The distance between 6th maxima and 10 th minima on the screen is closest to
- A \(12.0 \mathrm{~mm}\)
- B \(10.5 \mathrm{~mm}\)
- C \(14.25 \mathrm{~mm}\)
- D \(8.75 \mathrm{~mm}\)
Answer & Solution
Correct Answer
(D) \(8.75 \mathrm{~mm}\)
Step-by-step Solution
Detailed explanation
Given parameters are slit separation \(d = 0.2 \mathrm{~mm} = 0.2 \times 10^{-3} \mathrm{~m}\), distance to screen \(D = 1 \mathrm{~m}\), and wavelength \(\lambda = 500 \mathrm{~nm} = 500 \times 10^{-9} \mathrm{~m}\). The position of the \(n\)-th maxima is given by…
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