In a double-slit experiment, the fifth maximum is 2.8 cm from the central maximum on a screen that is 1.5 m away from the slits. If the slits are 0.15 mm apart, what is the wavelength of the light being used?

Explanation:

Screen distance D = 1.5 m

slit separation d = .15 x 10⁻³ m

distance of 5 th maxima   = 2.8 x 10⁻² m

distance of n th maxima = n x λ D/d

2.8 x 10⁻² = 5 x ( λ 1.5 / .15 x 10⁻³ )

λ  = (2.8 x 10⁻² x .15 x 10⁻³ ) / (5 x 1.5 )

= .42 x 10⁻⁵ / 7.5

.056 x 10⁻⁵

= 56 x 10⁻⁸ m

= 5600 A

The wavelength of the light is 560 nm.

Explanation:

Given that,

Distance from the central maximum y= 2.8 cm

Distance D=1.5 m

Distance between the slit = 0.15 m

We need to calculate the wavelength of the light

Using formula of constructive interference

Put the value into the formula

Hence, The wavelength of the light is 560 nm.

b

explanation:

fgravity=10x10=100n case 1

fgravity=30x20=600n case 2

6 times more