Q. Two slits separated by a distance of 1 mm are illuminated with red light of wavelength 6.5$\times 10^{- 7}$ m. the interference fringes are observed on a screen place 1 m from the slits. The distance between the third dark fringe and the fifth bright fringe is equal to
A) 0.65 mm
B) 1.63 mm
C) 3.25 mm
D) 4.88 mm
Explanation
(b) Here, $d = 1\ mm = 10^{- 3}m,$
$\lambda = 6.5 \times 10^{- 7}m$
$D = 1m$
$x_{5} = n\lambda\frac{D}{d} = 5 \times 6.5 \times 10^{- 7} \times \frac{1}{10^{- 3}}$
$\ \ \ \ \ = {32.5 \times 10}^{- 4}m$
$x_{3} = (2n - 1)\frac{\lambda}{2}\frac{D}{d}$
$\ \ \ \ = \frac{(2 \times 3 - 1) \times 6.5 \times 10^{- 7}}{2 \times 10^{- 3}}$
$= 16.25 \times 10^{- 4}\ m$
$x_{5} - x_{3} = (32.5 - 16.25)10^{- 4}m$
$= 16.25 \times 10^{- 4}m = 1.63\ mm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.