# A laser beam, with a wavelength of 532 nm, is directed exactly perpendicular to a screen having...

## Question:

A laser beam, with a wavelength of 532 nm, is directed exactly perpendicular to a screen having two narrow slits spaced 0.15 mm apart. Interference fringes, including a central maximum, are observed on a screen 2.5 m away. The direction of the beam is then slowly rotated around an axis parallel to the slits to an angle of 1.5 degrees. By what distance does the central maximum on the screen move?

## Interference:

The phenomenon in which two different waves initiated from two coherent sources meet each other, combine and form a new wave that may have higher or lower amplitude is known as interference.

Given Data

• The wavelength of laser beam is: {eq}\lambda = 532\;{\rm{nm}} {/eq}
• The distance of interference fringe from screen is: {eq}D = 2.5\;{\rm{mm}} {/eq}
• The rotation in the angle is: {eq}\phi = 1.5^\circ {/eq}

Calculate the distance moved by the central maximum on the screen.

{eq}\begin{align*} \tan \phi &= \dfrac{x}{D}\\ x &= D\tan \phi \end{align*} {/eq}

Substitute all the values in the above equation.

{eq}\begin{align*} x &= \left( {2.5\;{\rm{m}}} \right)\tan 1.5^\circ \\ x &= 0.06546\;{\rm{m}}\\ x &= 0.06546\;{\rm{m}}\dfrac{{{{10}^3}\;{\rm{mm}}}}{{1\;{\rm{m}}}}\\ x &= 65.46\;{\rm{mm}} \end{align*} {/eq}

Thus, the distance moved by the central maximum on the screen is {eq}65.46\;{\rm{mm}} {/eq}. 