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......?

Question

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......?

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2 answers
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Answer 1

Bot answered
1 year ago

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.

Answer 2

Step-by-Step Bot answered
1 year ago

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.