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o-na [289]
3 years ago
13

Monochromatic (single-wavelength) light falling on two slits 0.016 mm apart produces the fifth-order bright spot at an 8.8◦angle

. What is the wavelength of the light used?
Physics
2 answers:
Elena L [17]3 years ago
8 0

Given Information:  

distance = d = 0.016 mm = 0.0016 cm = 0.000016 m

Angle = θ = 8.8°

bright spot = m = 5

Required Information:  

Wavelength =  λ = ?

Answer:

Wavelength = 4.89x10⁻⁷ m

Explanation:

Monochromatic light passes through a double slit. The corresponding diffraction is given by the equation:

dsinθ = mλ

where d is the distance between two slits, m is the order of the diffraction, θ is the angle and λ is the wavelength of the light wave.

λ = dsinθ/m

λ = 0.016*sin(8.8°)/5

λ = 4.89x10⁻⁷ m

Therefore, the wavelength of the monochromatic light falling on two slits is 4.89x10⁻⁷ m

dimaraw [331]3 years ago
3 0

Answer:

the wavelength of the light used is 4.8 × 10⁻⁷m

Explanation:

We have given distance between the two slits

d=1.6\times 10^{-2}mm=1.6\times 10^{-5}m

Angle \Theta =8.8^{\circ}

We know that for nth order fringe

dsin\Theta =n\lambda

In question it is given that 5th order so n =5

So 1.6\times 10^{-5}sin\ 8.8 =5\lambda\\\lambda = 4.8\times 10^{-7}m

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