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RSB [31]
4 years ago
12

A diffraction grating has 2605 lines per centimeter, and it produces a principal maximum at = 30.3°. The grating is used with li

ght that contains all wavelengths between 410 and 660 nm. What are the two wavelengths of the incident light that could have produced this maximum? Give the longer wavelength as answer (a).
Physics
1 answer:
Yuri [45]4 years ago
6 0

Answer:

so m = 3 ,4 wavelength between 410 to 660 nm  so 484.25 nm and 645.6 nm light produce principal maximum

Explanation:

given data

grating =2605 lines per centimeter

angle = 30.3°

wavelengths between 410 and 660 nm

to find out

two wavelengths of the incident light that could have produced this maximum

solution

we know diffraction grating has 2605 lines / cm

so d = ( 1/ 2605 cm )

and we know equation

d sinθ = m× λ

so λ = d sinθ / m

θ = 30.3 so sin30.3 =

λ = (1/2605) sin30.3 / m

λ = (1937 nm ) / m

here put m = 1 , 2 , 3 , 4

if m = 1

λ = (1937 nm ) / 1  = 1937 nm

if m = 2

λ = (1937 nm ) / 2  = 968.5 nm

if m = 3

λ = (1937 nm ) / 3 = 645.6 nm

if m = 4

λ = (1937 nm ) / 4 = 484.25 nm

so that m = 3 ,4 wavelength between 410 to 660 nm  so 484.25 nm and 645.6 nm light produce principal maximum

and longest wavelength is between 410 to 660 nm is 645.6 nm

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Answer:

the block reaches higher than the sphere

\frac{y_{sphere}} {y_block} = 5/7

Explanation:

We are going to solve this interesting problem

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Let's use the concept of conservation of energy

starting point. At the top of the ramp

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final point. At the exit of the ramp

         Em_f = K + U = ½ m v² + ½ I w² + m g y₂

notice that we include the translational and rotational energy, we assume that the height of the exit ramp is y₂

energy is conserved

          Em₀ = Em_f

         m g y₁ = ½ m v² + ½ I w² + m g y₂

angular and linear velocity are related

        v = w r

the moment of inertia of a sphere is

         I = \frac{2}{5} m r²

we substitute

         m g (y₁ - y₂) = ½ m v² + ½ (\frac{2}{5} m r²) (\frac{v}{r})²

         m g h = ½ m v² (1 + \frac{2}{5})

where h is the difference in height between the two sides of the ramp

h = y₂ -y₁

         mg h = \frac{7}{5} (\frac{1}{2} m v²)

         v = √5/7  √2gh

This is the exit velocity of the vertical movement of the sphere

         v_sphere = 0.845 √2gh

B) is the same case, but for a box without friction

   starting point

          Em₀ = U = mg y₁

   final point

          Em_f = K + U = ½ m v² + m g y₂

          Em₀ = Em_f

          mg y₁ = ½ m v² + m g y₂

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         \frac{y_{sphere}} {y_bolck} = 5/7

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