Question

If in Young's double slit experiment, the distance between two slits is twice the wavelength of light used. If distance between slits and screen is $D$ then distance of first maxima from central maxima is -

• A. $D$
• B. $\frac{D}{2}$
• C. $\frac{D}{\sqrt{3}}$
• D. $\sqrt{3D}$

Answer 1

The correct option is C $\frac{D}{\sqrt{3}}$

The corresponding path difference,

$\Delta x=d\sin \theta =n\lambda$

$\Rightarrow d\sin \theta =\lambda \left[\text{For first maxima}\right]$

$\Rightarrow 2\lambda \sin \theta =\lambda$

$\Rightarrow \sin \theta =\frac{1}{2}$

$\Rightarrow \theta ={30}^{\circ }$

Now, $\tan {30}^{\circ }=\frac{y}{D}$

$\Rightarrow y=\frac{D}{\sqrt{3}}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.