Question

In Young's double slit experiment the light emited from source has $\lambda=6500 \,\mathring{A}$ and the distance between the two slits is $1 \,mm$. Distance between the screen and slit is 1 metre. Distance between third dark and fifth birth fringe will be :

• A. $3.2\, mm$
• B. $1.63 \,mm$
• C. $0.585\, mm$
• D. $2.31\, mm$

Answer 1

Answer: (B)

Distance of fifth bright fringe is given as
$x_{5 B }=n \frac{\lambda D}{d}=\frac{5 \times 6.5 \times 10^{-7} \times 1}{10^{-3}}$
$=32.5 \times 10^{-4} \,m$
Distance of third dark fringe is given as
$ x_{3 D }=(2 n-1) \frac{\lambda}{2} \frac{D}{d}=\frac{5 \times 6.3 \times 10^{-7} \times 1}{2 \times 10^{-3}}$
$=16.25 \times 10^{-4} m $
$\Rightarrow x_{5 B }-x_{3 D } \approx 1.63\, mm$ .