Question

help !! white light passes through a diffraction grating of 620 line per mm . First order visible spectrum appear on the wall 50 cm away . If wavelength of white light ranges from 400 nm to 700 nm , how wide is the first order spectrum ?

Answer 1

@JFraser can you help me with this one ??

Answer 2

i don't remember this part of spectra, sorry

but i'm sure there's an equation somewhere that might help. I like hyperphysics, they do a lot of math, but the diagrams are often very helpful

http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/grating.html

Answer 3

okay that does help a lot , but , the wavelength is given in range .. what should I do about that ?

Answer 4

Apply the equation for the largest value and the smallest value (700/400) thus you got the range.

Answer 5

umm , the range is 1.75 nm ?

Answer 6

\[d \sin \theta = n \lambda\]
d = 10 ^-3/620 ( see that link if you don't know why? http://openstudy.com/study#/updates/53f8cb8ee4b0ae4669f5d505 )

Answer 7

okay , last answer is 9.3 cm ? right ?

Answer 8

You will get theta1 and theta 2 at n=1
Then the width = 0.5 (tan theta2 - tan theta1).

Answer 9

wait if we're using the formula above , what is 50 cm refers to ? is it D or something else ?

Answer 10

Yea D

Answer 11

then , why are we using the d sin theta = n lambda

Answer 12

final answer is 0.11257 meter

Answer 13

to get theta

Answer 14

not y = ( m*lambda*D) / d

Answer 15

no

Answer 16

OK your question now is what is the equation for width right?

Answer 17

yes , I'm stuck with the 50 cm and the range wavelength . can't relate those with the question

Answer 18

\[\tan \theta = \frac{ y }{ D }\]
\[y _{1}= D \tan \theta _{1}.\]
\[y _{2}= D \tan \theta _{2}.\]
\[width = y _{2} - y _{1} = D \tan \theta _{2} - D \tan \theta _{1} = D ( \tan \theta _{2} - \tan \theta _{1})\]