Question

In a Young’s interference experiment, the two slits are separated by 0.150 mm and the incident light includes two wavelengths: l1 5 540 nm (green) and l2 5 450 nm (blue). The overlapping interference patterns are observed on a screen 1.40 m from the slits. Calculate the minimum distance from the center of the screen to a point where a bright fringe of the green light coincides with a bright fringe of the blue light:_____

Answer 1

Answer:

2.52 × 10⁻² cm

Explanation:

The distance of bright fringe from the center of the screen is given by the formula

                $y = \frac{m\lambda D}{d}$

Here, wavelength is λ, Distance of the screen from the slits is D, seperation between the

slits is d.

   

Separation between the slits, d = 0.15 mm

                                                     $= 0.15 * 10^{-3} m$

Distance of the screen from the slits = 1.40 m

We have a wavelength, λ1 = 540 nm

                                           = $540 * 10^{-9} m$

By substituting all these values in the above equation we get

                y1 = mλD/d

                $y1 = m(540 \times 10^{-9} m)(1.40 m)/(0.15 \times 10^{-3} m)$

                $y1 = m(5.04 * 10^{-3} m)$

We have a wavelength, λ2 = 450 nm

                                           = $450 * 10^-9 m$

By substituting all these values in the above equation we get

                $y_2 = \frac{m\lambda D}{d}$

                $y_2 = m(450 * 10^{-9} m)(1.40 m)/(0.15 * 10^{-3} m)$

                $y_2 = m'(4.20 * 10^{-3} m)$

According to the problem, these two distance are coincides with each other.

So,

                           $y_1 = y_2$

$m(5.04 * 10^{-3} m) = m'(4.20 * 10^{-3} m)$

by testing values, the above equation is satisfied only when, m = 5 and m' = 6

Then from the above we have

                           y1 = y2 = 0.0252 m

                                         = 2.52 × 10⁻² cm