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makvit [3.9K]
3 years ago
9

The second minimum in the diffraction pattern of a 0.11-mm-wide slit occurs at 0.72°. What is the wavelength of the light?

Physics
1 answer:
Anit [1.1K]3 years ago
6 0

Answer:

691.13 nm

Explanation:

d = width of the slit = 0.11 x 10⁻³ m

θ = angle of diffraction pattern = 0.72° degree

λ = wavelength of the light = ?

m = order = 2                              (since second minimum)

for the second minimum diffraction pattern we use the equation

d Sinθ = m λ

Inserting the values

(0.11 x 10⁻³) Sin0.72 = (2) λ

λ = 691.13 x 10⁻⁹ m

λ = 691.13 nm

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The Magnifying power of a telescope is M = 109.26

Explanation:

Radius of curvature R = 5.9 m = 590 cm

focal length of objective f_{objective} = \frac{R}{2}

⇒ f_{objective} = \frac{590}{2}

⇒ f_{objective} = 295 cm

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Magnifying power of a telescope is given by,

M = \frac{f_{objective} }{f_{eyepiece} }

M = \frac{295}{2.7}

M = 109.26

therefore the Magnifying power of a telescope is M = 109.26

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To determine the number of gold bars that would fit in the vault, determine the volume (v) of a single gold bar.
                              v = l x h x w
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