23P1Q16

16 D

$\displaystyle (\sin {{\theta }_{1}}=\frac{\lambda }{b})$

Assume small diffraction angles, so first minimal angle can be written as $\displaystyle {{\theta }_{1}}=\frac{\lambda }{b}$ .

In addition, width of the central maximum x can be written as the arc length subtended by angle $\displaystyle 2{{\theta }_{1}}$ .

In other words, $\displaystyle x=D\times 2{{\theta }_{1}}$

Substitute $\displaystyle {{\theta }_{1}}=\frac{\lambda }{b}$ : $\displaystyle \begin{aligned} & x=D\times 2\frac{\lambda }{b} \\ & bx=D\times 2\frac{c}{f} \\ & b=\frac{2cD}{fx} \end{aligned}$

Uncategorized

xmPhysics

23P1Q16

2 thoughts on “23P1Q16”

Leave a comment Cancel reply

Share this:

Related

2 thoughts on “23P1Q16”

Leave a comment Cancel reply