The distance between the 1st and 2nd maxima is 0.3 cm when the screen is located 2 m away. if d = 0.4 mm, a) what is the wavelength?
b) what is the size of central peak?

a) The wavelength of light is 6×10^-7 m.

b) The size of the central peak is 6×10^-3 m.

Explanation:

let:

α be the distance between the 1st and 2nd maxima

d = 0.4 mm be the grating slit space

X be the distance to the screen

a) α = λ×X/d

   λ = α×d/X

      = (0.3×10^-2)×(0.4×10^-3)/(2)

      = 6×10^-7 m

Therefore, the wavelength of light is 6×10^-7 m.

b) the size of the central peak is given by:

 D = 2×λ×X/d

     = 2×(6×10^-7)×(2)/(0.4×10^-3)

     = 6×10^-3 m

Therefore, the size of the central peak is 6×10^-3 m.

i think the answer is b

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