First-Order Spectrum Width for Diffraction Grating

[lovelylm1980]

White light containing wavelengths from 405 nm to 760 nm falls on a grating with 7510 lines/ cm. How wide is the first-order spectrum on a screen 2.29 m away?

Is this correct

sin theta= (1)*(4.05e-7m)/(1.33e-6m)= 0.305 theta=17.8
location= 2x(tan 17.8)= (2*2.29)*0.321=1.47m

sin theta= (1)*(7.60e-7m)/(1.33e-6m)= 0.571 theta=34.8
location= 2x(tan 34.8)= (2*2.29)*0.695= 3.18m

so then should the answer to the question be
3.18m+1.47m/2= 2.33m

 

[Doc Al]

location= 2x(tan 17.8)= (2*2.29)*0.321=1.47m

Why did you multiply by 2?

 

[lovelylm1980]

so then should i multiply the angle by 2

 

[Doc Al]

Why are you multiplying anything by 2??

 

[lovelylm1980]

because the book states that with large angles you should multiply by two either way I don't get the correct answer based on the formula i used. Whether I multiply by 2 or not the answer still seems to be wrong. Am I not supposed to add at the end then divide by two for the answer?

 

[Doc Al]

because the book states that with large angles you should multiply by two either way I don't get the correct answer based on the formula i used. Whether I multiply by 2 or not the answer still seems to be wrong. Am I not supposed to add at the end then divide by two for the answer?

That makes no sense.

Find the location of the first order maxima for each wavelength. Subtract those two numbers and you'll have the width of the first order spectrum.