A red laser from the physics lab is marked as producing 632.8-nm light. When light from this laser falls on two closely spaced slits, an interference pattern formed on a wall several meters away has bright red fringes spaced 5.50 mm apart near the center of the pattern. When the laser is replaced by a small laser pointer, the fringes are 5.62 mm apart. What is the wavelength of light produced by the pointer?

Question

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valetta valetta · Answer 1 · 2019-08-30T07:42:42+02:00

Answer:

646.60 nm

Step-by-step explanation:

Given:

wavelength of the light produced, λ₁ = 632.8 nm

Spacing between the bright fringes, y₁ = 5.50 mm

Spacing between the bright fringes when the laser is replaced,y₂ = 5.62 nm

Now,

the spacing between the bright fringes (y) is given as:

y = [tex]wavelength\times \frac{D}{d}[/tex]

where,

D is the distance between the slit and the screen

d is the distance between the slits

or

[tex]\frac{D}{d} = \frac{y}{\lambda}[/tex]

therefore, using the above relation, we have

[tex]\frac{y_1}{\lambda_1}=\frac{y_2}{\lambda_2}[/tex]

where,

λ₂ is the wavelength of the light produced by the smaller pointer

thus, on substituting the values,we get

[tex]\frac{5.50}{632.8}=\frac{5.62}{\lambda_2}[/tex]

or

λ₂ = 646.60 nm

Hence,

the wavelength of light produced by the smaller pointer = 646.60 nm