Question

In a Young's double slit experiment, the separation between the slits is $0.15\text{mm}$. In the experiment, a source of light of wavelength $589\text{nm}$ is used, and the interference pattern is observed on a screen kept $1.5\text{m}$ away. The separation between the successive bright fringes on the screen is:

• A. $6.9\text{mm}$
• B. $3.9\text{mm}$
• C. $5.9\text{mm}$
• D. $4.9\text{mm}$

Answer 1

The correct option is C $5.9\text{mm}$

Given that, the distance between screen and slits, $D=1.5\text{m}$
Separation between slits, $d=0.15\text{mm}$
Wavelength of source of light, $\lambda =589\text{nm}$

The separation between the successive bright fringe on the screen is Fringe width,

$\beta =\frac{D}{d}\lambda =\frac{1.5}{0.15\times {10}^{-3}}\times 589\times {10}^{-9}m$

$\beta =589\times {10}^{-5}=5.89\times {10}^{-3}\text{m}$

$\beta =5.89\text{mm}\approx 5.9\text{mm}$

Hence, option $\left(C\right)$ is correct.