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7. A blue laser with a wavelength of 480 nm and a yellow laser with a...

7. A blue laser with a wavelength of 480 nm and a yellow laser with a wave length of 550 nm illuminates a diffraction grating with 100 rulings per mm. A screen 2 m wide is placed 2 m from the diffraction grating. A. How many maxima are observed on the screen from the blue laser? B. What is the distance between the first maxima of the blue and green light?

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Answer #1

a)let the number of maxima to the bottom of the central maxima be m.

so,

w*sin(theta) = m*lambda

where w is the length of the slits and m is the order of the maxima. since the length of the screen is 2meters,

it is placed 1 meters on each sides.

so the maximum value of theta is theta=arctan(1/2)

=26.56 degrees

so,

wsin(theta)=m*lambda

or 10^-5 * 0.447 = m*480*10^-9

or m=9.3125

taking the integer part = 9

so total number of maxima = 9(one side) + 9 (other side) + 1(central maxima)

=19 maximas

b)

so distance = 1*(2/(10^-5))*(550-480)*10^-9

=0.014 m

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