Q . In Young’s experiment interference bands are produced on the screen placed at 1.5m from the two slits 0.15mm apart and illuminated by light of wavelength 6000Å. If the screen is now taken away from the slit by 50cm the change in the fringe width will be
(a) 2 ×10-4 m
(b) 2 ×10-3 m
(c) 6 ×10-3 m
(d) 7 ×10-3 m
Ans : (b)
Sol:
$\displaystyle \beta = \frac{\lambda D}{d} $ …(i)
$\displaystyle \beta’ = \frac{\lambda D’}{d}$ …(ii)
Where , D = 1.5 m , D’ = 1.5 + 0.5 = 2 m
d = 0.15 mm
∆β= β’-β