Question

Light of wavelength $520\text{nm}$ passing through a double slit, produces interference pattern of relative intensity versus deflection angle $\theta$ as shown in the figure. Find the separation $d$ between the slits.

• A. $1.98\times {10}^{-4}\text{m}$
• B. $1.98\times {10}^{-5}\text{m}$
• C. $1.98\times {10}^{-6}\text{m}$
• D. $1.98\times {10}^{-7}\text{m}$

Answer 1

The correct option is B $1.98\times {10}^{-5}\text{m}$

The first minimum is at an angle $\theta ={0.75}^{\circ }$

The path difference for $1^{st}$ minimum,

$\Delta x=d\sin \theta =d\theta =\frac{\lambda }{2}[\because \theta \text{is very small}]$

$\Rightarrow d\times 0.75\times \frac{\pi }{180}=\frac{520\times {10}^{-9}}{2}$

$\Rightarrow d=\frac{180\times 520\times {10}^{-9}}{2\times 3.14\times 0.75}$

$\Rightarrow d=1.98\times {10}^{-5}\text{m}$

Hence, option $\left(\text{B}\right)$ is correct.