Question

$\text{In YDSE,}D=1\text{m},d=1\text{mm}\text{and}\lambda =\frac{1}{2}\text{nm}$ the distance between the first and central maxima on the screen is-

• A. $\frac{1}{2}\text{m}$
• B. $\frac{1}{\sqrt{3}}\text{m}$
• C. $1\text{m}$
• D. None of the above

Answer 1

The correct option is B $\frac{1}{\sqrt{3}}\text{m}$

Given, $D=1\text{m},d=1\text{mm}\text{and}\lambda =\frac{1}{2}\text{nm}$


$\because D>>d$

Hence, path difference at any angular position $\theta$ on the screen is,

Hence, $\Delta x=d\sin \theta$

For first maxima,

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

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

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

From, the figure

$\theta =\frac{y}{D}$

$\Rightarrow y=D\tan \theta =\frac{1}{\sqrt{3}}\text{m}$

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