Question

A physics instructor wants to project a spectrum of visible-light colors from 400 nm to 700 nm as part of a classroom demonstration. She shines a beam of white light through a diffraction grating that has 700 lines per mm, projecting a pattern on a screen 2.9 m behind the grating. How wide is the spectrum corresponding to m=1?

Answer 1

Answer:

Explanation:

Given

$Slit\ width=d=\frac{10^{-3}}{700}=1.428\times 10^{-6} m$

Distance of screen from light source $D=2.9 m$

$\lambda _1=400 nm$

$\lambda _2=700 nm$

we know in slit Experiment

$d\sin \theta=n\lambda$, where $\theta$=angle between the path and a line from the slits to the screen

$1.428\times 10^{-6}\cdot \sin \theta =1\times 400\times 10^{-9}$

$\sin \theta =0.28011$

$\theta =16.26^{\circ}$

and $\tan \theta =\frac{y}{D}$

where y is the position of First maxima corresponding $\lambda _1$

$y_1=D\tan \theta$

$y_1=0.845 m$

calculations for  $\lambda _2$

$d\sin \theta=n\lambda _2$

$1.428\times 10^{-6}\cdot \sin \theta =1\times 700\times 10^{-9}$

$\sin \theta '=0.4901$

$\theta '=29.34^{\circ}$

$\tan \theta '=\frac{y'}{D}$

for n=1

$y_1'=D\tan \theta '$

$y_1'=2.9\times 0.5622=1.63 m$

Spectrum Width $=y_1'-y_1$

$=0.785 m$