JEE Mains · Physics · STD 12 - Wave optics
- JEE Mains 2026
- Easy
- Numerical
A beam of light consisting of wavelengths 650 nm and 550 nm illuminates the Young's double slits with separation of 2 mm such that the interference fringes are formed on a screen, placed at a distance of 1.2 m from the slits. The least distance of a point from the central maximum, where the bright fringes due to both the wavelengths coincide, is ___________ \(\times 10^{-5} m\).
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) 429
Step-by-step Solution
Detailed explanation
\(y=n\frac{\lambda D}{d}\)
\(y_{1}=y_{2}\)
\(n_{1}\lambda_{1}\frac{D}{d}=n_{2}\lambda_{2}\frac{D}{d}\)
\(\frac{n_{1}}{n_{2}}=\frac{\lambda_{2}}{\lambda_{1}}=\frac{550}{650}=\frac{11}{13}\)
\(y=11\times\frac{\lambda_{1}D}{d}=\frac{11\times650\times10^{-9}\times1.2}{2\times10^{-3}}\)
\(y=429\times10^{-5}\)
\(y_{1}=y_{2}\)
\(n_{1}\lambda_{1}\frac{D}{d}=n_{2}\lambda_{2}\frac{D}{d}\)
\(\frac{n_{1}}{n_{2}}=\frac{\lambda_{2}}{\lambda_{1}}=\frac{550}{650}=\frac{11}{13}\)
\(y=11\times\frac{\lambda_{1}D}{d}=\frac{11\times650\times10^{-9}\times1.2}{2\times10^{-3}}\)
\(y=429\times10^{-5}\)
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