Question

In Young’s double-slit experiment, the distance between two sources is $0.1mm$. The distance of the screen from the sources is $20cm$. The wavelength of light used is $5460\AA$ Then the angular position of the first dark fringe is$A$.

• A. $0.{08}^o$
• B. $0.{16}^o$
• C. $0.{20}^o$
• D. $0.{32}^o$

Answer 1

The correct option is B

$0.{16}^o$

Solution:

Step 1: Given information

The distance between two sources$=0.1mm$

The distance of the screen from the sources$=20cm$

The wavelength of light used$=5460\AA$

Step 2: Prove the angular position of the first dark fringe is$A$

Knows $\Delta x=d\sin \theta$, and $\Delta x=\frac{\lambda }{2}$

Therefore,

$\Rightarrow \sin \theta =\frac{\lambda }{2d}$

$\Rightarrow \theta ={\sin }^{-1}\left(\frac{\lambda }{2d}\right)$

$\Rightarrow \theta ={\sin }^{-1}\left[\frac{(5460x{10}^{-10})}{(2x0.1x{10}^{-3})}\right]$

$\Rightarrow \theta =0.{16}^o$

Hence, the correct option is (B).