Optics Question 846
Question: 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 placed 1 m from the silts. The distance of the third dark fringe from the central fringe will be equal to:
Options:
A) 0.65 mm
B) 1.30mm
C) 1.62 mm
D) 1.95mm
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Given $ d=1mm,\lambda =6.5\times {{10}^{-7}}m,D=1m $ .
For $ n^{th} $ dark fringe $ y _{n}=( \frac{2n-1}{2} )\frac{D\lambda }{d} $ .
For third fringe, $ y _{3}=( \frac{2\times 3-1}{2} )\times \frac{1\times 6.5\times {{10}^{-7}}}{1\times {{10}^{-3}}}=1.625mm $ .
BETA
