Question

In a double slit experiment, green light of $5303\stackrel{\circ }{A}$ falls on a double slit having a separation of $19.44\mu m$ and a width of $4.05\mu m$. The number of bright fringes between the first and the second diffraction minima is :

• A. $9$
• B. $10$
• C. $6$
• D. $5$

Answer 1

The correct option is D $5$

First minima:
$\sin {\theta }_{\text{dark}}=\frac{\lambda }{a}$
$y_1=\frac{\lambda D}{a}$

Second minima:
$y_2=\frac{2\lambda D}{a}$

Optical Path difference:
$\Delta x_n=\frac{d{y}_n}{D}$

$\begin{array}{cc}{y}_1\parallel \Delta x_1& =\frac{\lambda D}{a}\times \frac{d}{D}\\ & =\frac{\lambda d}{a}=\frac{19.44}{4.05}\lambda \\ & \sim 4.8\lambda \end{array}$

$\begin{array}{cc}{y}_2\parallel \Delta x_2& =\frac{2\lambda D}{a}\times \frac{d}{D}\\ & =\frac{2\lambda d}{a}=\frac{2\times 19.44}{4.05}\lambda \\ & \sim 9.6\lambda \end{array}$

Between $4.8\lambda$ and $9.6\lambda$… 5 bright fringes are formed.