In Young's double slit experiment carries out with light of wavelength =5000Angstrom, the distance between the slit us 0.2mm and the screen is at 200cm from the slit. The central maximum is at x=0.the third maximum (taking the central maximum as zeroth maximum) will be at x equal to......?

Answer 1

The distance between the central maximum and the first minimum is given by:

d = λD/d

where λ is the wavelength of light, D is the distance between the slits and the screen, and d is the distance between the two slits.

Plugging in the values, we get:

d = (5000 × 10^-10)m × 200m / 0.0002m
d = 0.1m

Therefore, the distance between successive maxima is d/2 = 0.05m.

The third maximum is located at x = 3d/2, or:

x = 3(0.05m) / 2
x = 0.075m

So the third maximum is located 7.5 cm away from the central maximum.