In a Young's double slit experiment, the separation between the slits is $0.15 mathrm{~mm}$.

Question:

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

  1. $6.9 \mathrm{~mm}$

  2. $3.9 \mathrm{~mm}$

  3. $5.9 \mathrm{~mm}$

  4. $4.9 \mathrm{~mm}$


Correct Option: , 3

Solution:

(3) Given, distance between screen and slits, $D=1.5 \mathrm{~m}$

Seperation between slits, $\mathrm{d}=0.15 \mathrm{~mm}$

Wavelength of source of light, $\lambda=589 \mathrm{~nm}$

Fringe-width, $w=\frac{D}{d} \lambda=\frac{1.5}{0.15 \times 10^{-3}} \times 589 \times 10^{-9} \mathrm{~m}$

$=589 \times 10^{-2} \mathrm{~mm}=5.89 \mathrm{~mm} \approx 5.9 \mathrm{~mm}$

Leave a comment

Close

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now