Question

Violet light is incident on a double slit. The distance between the two slits is 0.040mm, and aninterference pattern is created ona screen that is 2.0m away from the slits. If the 5th bright fringe on the screen is 10.0cm away from the central fringe, what is the wavelength of the light?

Answer 1

Answer:

The wavelength of the light is $0.04 \times 10^{-5}$ m

Explanation:

Given :

Distance between two slit $d = 0.040 \times 10^{-3}$ m

Distance of screen $D = 2$ m

Distance between central fringe to 5th bright fringe, $x = 10 \times 10^{-2}$ m

From the formula of interference,

    $d \sin \theta = n \lambda$

Where $\sin \theta$ ≅ $\theta$ we put $\theta = \frac{x}{D}$

   $\frac{dx}{D} = n\lambda$

Where $n =$ 5                             ( given in question )

Now we have to find wavelength of the light,

    $\lambda = \frac{dx}{nD}$

    $\lambda = \frac{0.040 \times 10^{-3} \times 10 \times 10^{-2} }{5 \times 2}$

    $\lambda = 0.04 \times 10^{-5}$ m

Therefore, the wavelength of the light is $0.04 \times 10^{-5}$ m