In a Young’s double slit experiment two slits are separated by
Question:

In a Young’s double slit experiment two slits are separated by $2 \mathrm{~mm}$ and the screen is placed one meter away. When a light of wavelength $500 \mathrm{~nm}$ is used, the fringe separation will be :

  1. $0.75 \mathrm{~mm}$

  2. $0.50 \mathrm{~mm}$

  3. $1 \mathrm{~mm}$

  4. $0.25 \mathrm{~mm}$


Correct Option: , 4

Solution:

(4)

Fringe width $(\beta)=\frac{\lambda D}{d}$

$d=2 \times 10^{-3} \mathrm{~m}$

$\lambda=500 \times 10^{-9} \mathrm{~m}$

$\mathrm{D}=1 \mathrm{~m}$

Now

$\beta=\frac{500 \times 10^{-9} \times 1}{2 \times 10^{-3}}$

$\beta=\frac{5}{2} \times 10^{-4}$

$\beta=2.5 \times 10^{-4}$

$\beta=0.25 \mathrm{~mm}$

Administrator

Leave a comment

Please enter comment.
Please enter your name.