Question

With Young's two slit arrangement, the distance between the slits is 1 mm. The distance between the slits and the screen is one metre. The two slits are equally illuminated with light of wavelength $6000\dot{A}$. Then the minimum distance from the central maximum to a point where the intensity is half of the maximum is:

• A. 0.6 mm
• B. 0.3 mm
• C. 0.15 mm
• D. 2.4 mm

Answer 1

The correct option is C 0.15 mm

d = 1 mm = 0.1 cm, D = 100 cm, $\lambda =6\times {10}^{-7}m$
The maximum intensity is $(\sqrt{I}+\sqrt{I}{)}^2=4I$
$\therefore 2I=I+I+2Icos\delta ,cos\delta =0,\delta =\frac{\pi }{2}$
Path difference = $\frac{\lambda }{2\pi }.\delta =\frac{\lambda }{2\pi }\frac{\pi }{2}=\frac{\lambda }{4}$
$\therefore \frac{dx}{D}=\frac{\lambda }{4}$ or $x=\frac{D\lambda }{4d}$ or $x=0.15mm$