Skip to content
dronacharya.ai
← JEE Main 2017 · 2 Apr · Shift 1
PhysicsSingle correct+4 / -1Question 23 of 90

JEE Main 2017 · 2 Apr · Shift 1, Physics question 23

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 :

Choose one option
Answer and worked solution

Official answer: B

The bright fringes coincide where n1(650)=n2(520)n_1(650)=n_2(520); the smallest solution is 4×650=5×520=26004\times650=5\times520=2600 nm.
y=2600×10−9×1.50.5×10−3=7.8 mm.y=\frac{2600\times10^{-9}\times1.5}{0.5\times10^{-3}}=7.8\text{ mm}.

Topics:Interference and Young's double slit
← Previous questionNext question →