Question

The lab experiment is repeated with the same two glass plates but with a different human hair. The sodium vapor lamp is replaced by a monochromatic light source that produces light of wavelength λ

air =590 nm. The separation between adjacent dark fringes is measured to be 0. 455 mm. Calculate the diameter of the hair. (The length of the plates is 12. 0 cm. Μm 2. [-/15 Points] TAMUCOLPHYSEML1 6. POST. 2. The lab experiment is repeated again with the same two glass plates and the same sodium lamp but with a different human hair. The space between the plates is filled with a transparent silicon grease that has n=1. 38. The glass plates each have refractive index 1. 20. The separation between adjacent dark fringes is measured to be 0. 780 mm. Calculate the diameter of the hair. (The wavelength emitted by the lamp is 589. 3 nm. The length of the plates is 12. 0 cm. )

Answer 1

Here are the step-by-step solutions for the two parts of this physics experiment question:

Part 1:

Given:

Wavelength of monochromatic light (λ) = 590 nm

Separation between adjacent dark fringes (d) = 0.455 mm

Length of glass plates (L) = 12.0 cm

Using the formula for interference fringe width:

d = λL/D

Where D is the diameter of the hair.

Rearranging and plugging in the given values:

D = λL/d

= (590 x 10-9 m)(0.12 m)/(0.455 x 10-3 m)

= 1.54 x 10-5 m

Therefore, the diameter of the hair is 1.54 x 10-5 m or 15.4 μm.

Part 2:

Given:

Wavelength of sodium lamp (λ) = 589.3 nm

Refractive index of glass plates (n1) = 1.20

Refractive index of silicon grease (n2) = 1.38

Separation between fringes (d) = 0.780 mm

Length of plates (L) = 12.0 cm

Using the formula for fringe width:

d = λL/(n1 - n2)D

Rearranging and plugging in values:

D = λL/(n1 - n2)d

= (589.3 x 10-9 m)(0.12 m)/[(1.20 - 1.38)(0.780 x 10-3 m)]

= 2.63 x 10-5 m

Therefore, the diameter of the hair is 2.63 x 10-5 m or 26.3 μm.