Wave Optics Problems: Diffraction Grating #1 & #2

[G01]

#1 A diffraction grating having 500lines/mm diffracts visible light at 30 degrees(\pi/6) What is its wavelength?

the distance between slits is .002mm(500l^-1)

d\sin\frac{\pi}{6} = \lambda = 1000nm, which is too big. The answer is 500nm. What am I doing wrong?

#2

The slit spacing in a diffration grating is .0002m. The screen is 2.0m behind the grating. The distance of the first bright fringe is .004m. What is the wavelength of light?

y = L\tan\theta
\tan^{-1}.004/2 = \theta = .0019
d\sin\theta = \lambda
.0002m\sin(.0019) = \lambda = 400nm

If this is right then I read from a graph wrong. If this is wrong, then I would appreciate anybody's input. Thank you for your time.

 

[nrqed]

#1 A diffraction grating having 500lines/mm diffracts visible light at 30 degrees(\pi/6) What is its wavelength?

the distance between slits is .002mm(500l^-1)

d\sin\frac{\pi}{6} = \lambda = 1000nm, which is too big. The answer is 500nm. What am I doing wrong?
.

The correct equation is

d\sin\frac{\pi}{6} = n \lambda

where is an integer. since n=1 gives you a result out of the visible spectrum (in the infrared), you try n=2, which gives something in the visible.


Pat

 

[G01]

AHHH icic, very simple now that I see it. Anybody for # 2

 

[G01]

bumping the thread...