Question

Two plane plates of glass are in contact in one edge and are separated at a point 20 cm from that edge by a wire of diameter 0.05 mm. What will be the width between the consecutive dark interference fringes formed when light of wavelength 589nm falls normally on air film enclosed between the plates.

Answer 1

As per the question,

distance between the glass plate (OA)=20cm

diameter of the wire (AB)=0.05mm

wavelength of light $(\lambda)=589nm$

angle $(\theta)=\frac{AB}{OA}=\frac{0.005cm}{20cm}=0.00025rad$

Fringe width $(\beta)=\frac{\lambda}{2\theta}$

$=\frac{589.3nm}{2\times 0.00025rad}$

$=1.1786mm$