JEE Main Solved Paper 2017 Question 20
Question: In a Young’s double slit experiment, slits are separated by 0.5 mm, and the screen is placed 150 cm away. A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide is: [JEE Main Solved Paper-2017]
Options:
A) 9.75 mm
B) 15.6 mm
C) 1.56 mm
D) 7.8 mm
Show Answer
Answer:
Correct Answer: D
Solution:
- For common maxima $ n _1{\lambda_1}=n _2{\lambda_2} $ $ n _1\times 650=n _2\times 520 $ $ \frac{n _1}{n _2}=\frac{4}{5} $ $ \frac{yd}{D}=n\lambda $ $ y=\frac{4\times 650\times {10^{-9}}\times 1.5}{0.5\times {10^{-3}}} $ $ y=7.8,mm $
