# In a double-slit experiment, the distance between the slits is 0.2 mm, and the distance to the...

## Question:

In a double-slit experiment, the distance between the slits is {eq}0.2\ mm {/eq}, and the distance to the screen is {eq}150\ cm {/eq}. What wavelength (in nm) is needed to have the intensity at a point {eq}1\ mm {/eq} from the central maximum on the screen be {eq}80\% {/eq} of the maximum intensity?

## Double Slit Experiment:

Thomas Young proposed the double-slit experiment to study the phenomenon of interference of light. It was observed that there were alternate bright and dark fringes that were formed on a screen when a monochromatic light was passed through a pair of slits whose width was equivalent to the wavelength of the light used. The intensity of the fringes kept on decreasing from the central maximum.

Given:

• Width of slits {eq}d = 0.2 mm = 0.2 \times 10^{-3} m {/eq}
• Distance of screen D = 150 cm = 1.5 m

In the ideal case, the intensity of each bright fringe is equal to the central maximum and they are equally spaced from each other.

But when it comes to reality, the intensity of the consecutive bright fringes varies and keeps on decreasing. If the intentiay of the sentral maximum is given by {eq}I_{max} {/eq}, then the intensity of a bright fringe at a distance y from the central maximum is given as:

{eq}I = I_{max} cos^2 (\dfrac{\pi d}{\lambda D} y) {/eq}

Given the intenisty at a distance {eq}y = 1 mm = 10^{-3} m {/eq} is 80% of the central maxiumum, then:

{eq}0.8 I_{max} = I_{max} cos^2 (\dfrac{\pi d}{\lambda D} y) {/eq}

{eq}0.8 =cos^2 (\dfrac{\pi d}{\lambda D} y) {/eq}

{eq}\sqrt{0.8} =cos (\dfrac{\pi d}{\lambda D} y) {/eq}

{eq}0.894 =cos (\dfrac{\pi d}{\lambda D} y) {/eq}

{eq}cos^{-1}(0.894 )= (\dfrac{\pi d}{\lambda D} y) {/eq}

{eq}0.465 = (\dfrac{\pi d}{\lambda D} y) {/eq}

{eq}\lambda = (\dfrac{\pi d}{(0.465) D} y) = (\dfrac{\pi (0.2 \times 10^{-3})}{(0.465) (1.5)} \times 10^{-3}) {/eq}

{eq}\lambda = 900 \times 10^{-9} m = 900 nm {/eq}