Question

https//esclet Table 2: Distance Measurements Measurement of L Grating Moasurement of X 1000 lines/mm PART 2 3. Observations of the laser pointer shone through the 1000 lines/mm diffraction grating: 4. Observations and drawings of the flashlight shone through the 1000 lines/mm diffraction grating: CALCULATIONS Calculate the wavelength of the laser for d 1/1000 mm. Use the modified "grating equation" given below. Remember to keep all calculations in millimeters (mm) until the end, and then convert your answer to nanometers (1 nm= 10-9 m ). Show all your work below:

Answer 1

Solution:

As we know;

nλ = dsinθ

i.e,

given d = 1/1000 mm => 10^-6 m or 1 micrometre

and

The n th-order maximum for each wavelength occurs at a specific angle.

All wavelengths are seen at θ =0, corresponding to n=0 as sinθ = 0 at 0 degrees.

next,the zeroth-order maximum (m=1) is observed at the angle that satisfies the relationship sin θ =λ/d

and the second-order maximum (m=2) is observed at a larger angle θ ....

Now using d= 10^-6 m

we get;

nλ = (10^-6) sinθ

and sinθ can take values between [0,1] as the angle takes values between: [0,90 ] degrees.

On an average:

nλ = (10^-6) (1/2)

=>   nλ = 5 x 10^-7 m

For n = 1,2,3,4 . . .

=> λ = 5 x 10^-7 m , 2.5 x 10^-7 m , 1.667 x 10^-7 m

or

λ = 500 nm , 250 nm , 167 nm ,and so on . . . .