Question

Monochromatic green light of wavelength 550 nm illuminates two parallel narrow slits 7.7$\mu$m apart. The angular deviation $\theta$ of third order (for m $=$ 3) bright fringe in radian and in degrees respectively are :

• A. $21.6,{12.4}^{\circ }$
• B. $0.216,{1.24}^{\circ }$
• C. $0.216,{12.4}^{\circ }$
• D. $216,{1.24}^{\circ }$

Answer 1

Answer: (C)

$0.216,{12.4}^{\circ }$

$dsin\theta =n\lambda$
$sin\theta =\frac{3\times 550\times {10}^{-9}}{7.7\times {10}^{-6}}$ (taking n=3)

$\simeq 216\times {10}^{-3}$
$=0.216$
(Here $\theta$ is small so we can take sin $\theta \approx \theta )$

$\theta =0.216$ radian
Now, $\pi radian={180}^0$
$\Rightarrow 3.14radian={180}^0$
So, $0.216radian=\frac{0.216}{3.14}\times {180}^0$$={12.4}^0$