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.

float.

archimedes' principle

denser than water sinks in water.

denser than mercury would sink in it