# At what angle will 630-nm light produce a second-order maximum when falling on a grating whose...

## Question:

At what angle will 630-nm light produce a second-order maximum when falling on a grating whose slits are {eq}1.15 \times 10^{-3} {/eq} cm apart? Use 3 significant figures.

## Angle made by second-order maximum

The angle of the second-order maximum or the second bright fringe after the central maximum can be found out using the relation:

{eq}\rm dsin(\theta) = m\lambda {/eq}

where d is the separation of the slits, m is the order of fringe and {eq}\lambda {/eq} is the wavelength of light.

The wavelength of light = {eq}\rm \lambda= 630 \ nm = 630 \times 10^{-9} \ m {/eq}

The separation of the slits = {eq}\rm d = 1.15 \times 10^{-3} \ cm =1.15 \times 10^{-5} \ m {/eq}

For second-order maximum, we have m = 2. Then the angle made by second-order maximum is:

{eq}\rm dsin(\theta) = m\lambda {/eq}

{eq}\rm 1.15 \times 10^{-5} sin(\theta) = 2 (630 \times 10^{-9}) {/eq}

{eq}\rm sin(\theta) = \dfrac{2 \times 630 \times 10^{-9}}{1.15 \times 10^{-5}} {/eq}

{eq}\rm sin(\theta) = 0.1095 {/eq}

{eq}\rm \theta = 6.3 ^{\circ} {/eq}