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GUJCET · Physics · Wave Optics
In young's double slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 12 cm . Then the wavelength of light used in the experiment is __________ .
- A 660 nm
- B 550 nm
- C 600 nm
- D 500 nm
Answer & Solution
Correct Answer
(C) 600 nm
Step-by-step Solution
Detailed explanation
(C) 600 nm
Slits separation \(d=0.28 \mathrm{~mm}\)
\(
\text { Screen } \begin{aligned}
\mathrm{D} & =1.4 \mathrm{~m} \\
n & =4^{\text {th }} \text { bright }
\end{aligned}
\)
Distance between the central bright fringe and the fourth bright fringe \(x=1.2 \mathrm{~cm}\)
For bright fringe path difference
\(\frac{x d}{\mathrm{D}}=n \lambda\)
\(\therefore \quad \lambda=\frac{x_{4} d}{n \mathrm{D}}\)
\(\therefore \lambda=\frac{1.2 \times 10^{-2} \times 0.28 \times 10^{-3}}{4 \times 1.4}\)
\(
\begin{aligned}
& \lambda=6 \times 10^{-7} \mathrm{~m} \\
& \lambda=600 \mathrm{~nm}
\end{aligned}
\)
Slits separation \(d=0.28 \mathrm{~mm}\)
\(
\text { Screen } \begin{aligned}
\mathrm{D} & =1.4 \mathrm{~m} \\
n & =4^{\text {th }} \text { bright }
\end{aligned}
\)
Distance between the central bright fringe and the fourth bright fringe \(x=1.2 \mathrm{~cm}\)
For bright fringe path difference
\(\frac{x d}{\mathrm{D}}=n \lambda\)
\(\therefore \quad \lambda=\frac{x_{4} d}{n \mathrm{D}}\)
\(\therefore \lambda=\frac{1.2 \times 10^{-2} \times 0.28 \times 10^{-3}}{4 \times 1.4}\)
\(
\begin{aligned}
& \lambda=6 \times 10^{-7} \mathrm{~m} \\
& \lambda=600 \mathrm{~nm}
\end{aligned}
\)
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