* JUST NEED HELP FOR d) THE REST WAS SOLVED SUCESSFULLY ALONE* A beam of light c
ID: 1462348 • Letter: #
Question
*JUST NEED HELP FOR d) THE REST WAS SOLVED SUCESSFULLY ALONE*
A beam of light contains all wavelengths between 420 nm and 680 nm and is incident on a diffraction grating with 170 lines/mm.
a) If two consecutive orders are overlapping, what can you say about the angles of the high and low wavelength light from the two orders?
answer: [420nm]higher order<[680nm]lower order
b) What is the lowest order that is overlapped by another order?
answer: m= 2
c) What is the highest order for which the complete wavelength range of the beam is present?
answer: m= 8
d) What is the greatest angle at which the light with the shortest wavelength appears?
_______ degrees
Explanation / Answer
find the highest n at which sin for 420 nm is less than 1
then calculating the angle
d sin = n
d = 1/170 mm/line ( 1E6 nm/mm) = 5882 nm/line
n = 0,1,2,3...n (order of image)
therefore, sin = n / d <1
(n 420/ 5882 ) <1
maximaum integer value of n= 14
For which sin = 1
Therefore, greatest angle at which the light with the shortest wavelength appears= 90 degrees
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