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Monochromatic light of wavelength 687 nm is incident on a narrow slit. On a screen 1.65 m away, the distance between the second diffraction minimum and the central maximum is 2.09 cm. (a) Calculate the angle of diffraction θ of the second minimum. (b) Find the width of the slit. Please explain with your solution

User Lyon
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Answer:

a ) 1.267 radian

b ) 1.084 10⁻³ mm

Step-by-step explanation:

Distance of screen D = 1.65 m

Width of slit d = ?

Wave length of light λ = 687 nm.

Distance of second minimum fro centre y = 2.09 cm

Angle of diffraction = y / D

= 2.09 /1.65

= 1.267. radian

Angle of diffraction of second minimum

= 2 λ / d

so 2 λ / d = 1.267

d = 2 λ / 1.267 = (2 x 687 ) /1.267 nm

=1084.45 nm = 1.084 x 10⁻³ mm.

User Allen Fernandes
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