Question

You have a double slit experiment, with the distance between the two slits to be 0.025 cm. A screern is 120 cm behind the double slits. The distance between the central maximum and the Sh maximum is 1.52 cm. Please calculate the wavelength of the light used in the experiment

Answer 1

Answer:

The wavelength of the light is 633 nm.

Explanation:

Given that,

Distance between the two slits d= 0.025 cm

Distance between the screen and slits D = 120 cm

Distance between the slits y= 1.52 cm

We need to calculate the angle

Using formula of double slit

$\tan\theta=\dfrac{y}{D}$

Where, y = Distance between the slits

D = Distance between the screen and slits

Put the value into the formula

$\tan\theta=\dfrac{1.52}{120}$

$\theta=\tan^{-1}\dfrac{1.52}{120}$

$\theta=0.725$

We need to calculate the wavelength

Using formula of wavelength

$d\sin\theta=n\lambda$

Put the value into the formula

$0.025\times\sin0.725=5\times\lambda$

$\lambda=\dfrac{0.025\times10^{-2}\times\sin0.725}{5}$

$\lambda=6.326\times10^{-7}\ m$

$\lambda=633\ nm$

Hence, The wavelength of the light is 633 nm.