99 Percentile Qs Bank for JEE MainPhysicsWave Optics
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:
Options
- A15 . 6 mm
- B1 . 56 mm
- C7 . 8 mm
- D9 . 75 mm
Correct answer
C. 7 . 8 mm
Step-by-step solution
y 1 = n 1 λ 1 D d for bright fringes y 2 = n 2 λ 2 D d for bright fringes To coincide n 1 λ 1 = n 2 λ 2 n 1 × 650 = n 2 × 520 ∴ n 1 n 2 = 4 5 For minimum value of n 1 and n 2 n 1 = 4 n 2 = 5 y 1 m i n = 4 × 650 × n m × 150 m 0.5 × 10 - 3 m = 7800 × 10 - 6 m = 7.8 m m