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kkurt [141]
3 years ago
8

You are comparing two diffraction gratings using two different lasers: a green laser and a red laser. You do these two experimen

ts
1. Shining the green laser through grating A you see the first maximum 1 meter away from the center
2. Shining the red laser through grating B, you see the first maximum 1 meter away from the center

In both cases, the gratings are the same distance from the screen.

(a) What can you deduce about the gratings?
(b) What would you observe if you shone the green laser through grating B:?

a. (a) grating A has more lines/mm; (b) the first maximum less than 1 meter away from the center
b. (a) grating B has more lines/mm; (b) the first maximum less than 1 meter away from the center
c.(a) grating B has more lines/mm; (b) the first maximum more than 1 meter away from the center
d. (a) grating B has more lines/mm: (b) the first maximum 1 meter away from the center
e. (a) grating A has more lines/mm; (b) the first maximum more than 1 meter away from the center
f. (a) grating A has more lines/mm; (b) the first maximum 1 meter away from the center
Physics
1 answer:
Nikolay [14]3 years ago
5 0

Answer:

a. (a) grating A has more lines/mm; (b) the first maximum less than 1 meter away from the center

Explanation:

Let  n₁ and n₂ be no of lines per unit length  of grating A and B respectively.

λ₁ and λ₂ be wave lengths of green and red respectively , D be distance of screen and d₁ and d₂ be distance between two slits of grating A and B ,

Distance of first maxima for green light

= λ₁ D/ d₁

Distance of first maxima for red light

= λ₂ D/ d₂

Given that

λ₁ D/ d₁ = λ₂ D/ d₂

λ₁ / d₁ = λ₂ / d₂

λ₁ / λ₂  = d₁ / d₂

But

λ₁  <  λ₂

d₁ < d₂

Therefore no of lines per unit length of grating A will be more because

no of lines per unit length  ∝ 1 / d

If grating B is illuminated with green light first maxima will be at distance

λ₁ D/ d₂

As λ₁ < λ₂

λ₁ D/ d₂ < λ₂ D/ d₂

λ₁ D/ d₂ < 1 m

In this case position of first maxima will be less than 1 meter.

Option a is correct .

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(a) 638.4 J

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Here we want to calculate the work done by the force F, of magnitude

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g=9.8 m/s^2 is the acceleration of gravity

Substituting,

W=(19.6)(9.8)=192.1 N

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d = 4.20 m

The gravitational force acts vertically downward, so the angle between the displacement and the force is

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Where \alpha = 24^{\circ} is the angle between the incline and the horizontal.

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The magnitude of the normal force is equal to the component of the weight perpendicular to the ramp, therefore:

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\alpha=24^{\circ}

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R=(19.6)(9.8)(cos 24)=175.5 N

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Therefore, the work done by the normal force is

W_R=R d cos \theta =(175.4)(4.20)(cos 90)=0

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