Question

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?

Answer 1

Step-by-step 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

Answer 2

The wavelength of the light is 560 nm.

Step-by-step 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

`m\lambda=(yd)/(D)`

`\lambda=(yd)/(mD)`

Put the value into the formula

`\lambda=(2.8**10^(-2)*0.15*10^(-3))/(5*1.5)`

`\lambda=5.6*10^(-7)`

`\lambda=560\ nm`

Hence, The wavelength of the light is 560 nm.