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......?
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.
To find the position of the third maximum in Young's double slit experiment, we can use the formula:
y = mλL / d
where:
y = position of the maximum
m = order of the maximum (0 for central maximum)
λ = wavelength of the light
L = distance between the slit and the screen
d = distance between the two slits
Given:
Wavelength (λ) = 5000 Angstrom = 5000 x 10^-10 meters
Distance between the slits (d) = 0.2 mm = 0.2 x 10^-3 meters
Distance between the slit and the screen (L) = 200 cm = 200 x 10^-2 meters
To find the position of the third maximum, we set m = 3:
y = (3 * 5000 x 10^-10 * 200 x 10^-2) / (0.2 x 10^-3)
Simplifying,
y = (3 * 5000 * 200) / 0.2
y = 3 * 5000000
y = 15000000
Therefore, the position of the third maximum (taking the central maximum as zeroth maximum) will be at x = 15,000,000.