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Hitman42 [59]
3 years ago
9

The radius of curvature of both sides of a converging lens is 18 cm. One side of the lens is coated withsilver so that the inner

surface is reflective. When light is incident on the uncoated side it passes throughthe lens, reflects off the silver coating, and passes back through the lens. The overall effect is that of amirror with focal length 5.0 cm. What is the index of refraction of the lens material
Physics
1 answer:
Dahasolnce [82]3 years ago
7 0

Answer:

n = 1.4

Explanation:

Given,

R1 = 18 cm, R2 = -18 cm

From lens makers formula

1/f = (n - 1)(1/18 + 1/18) = (n-1)/9

f = 9/(n-1)

Power, P = 1/f ( in m) = (n-1)/0.09

Now, this lens is in with conjunction with a concave mirror which then can be thought of as to be in conjunction with another thin lens

Power of concave mirror = P' = 1/f ( in m) = 2/R = 2/0.18 = 1/0.09

Net power of the combination = 2P + P' = 2(n-1)/0.09 + 1/0.09 = 1/0.05

n = 1.4

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Answer:

Distance = 16.9 m

Explanation:

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Imagine you are holding a box in your hand, the force required for you to hold the box with mass 4.54kg and you want to accelera
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Answer:

(a) Fₓ = 0 N

F_y = 9.08 N

(b)

(a) Fₓ = 0 N

<u></u>F_y = 9.08 N

(c) F_y = 0 N

Fₓ = 9.08 N

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F = force = ?

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Therefore,

F = (4.54 kg)(2 m/s²)

F = 9.08 N

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(a)

Since the motion is in the vertical direction. Therefore the magnitude of the force in x-direction will be zero:

<u>Fₓ = 0 N</u>

For upward direction the force will be positive:

<u></u>F_y<u> = 9.08 N</u>

<u></u>

(b)

Since the motion is in the vertical direction. Therefore the magnitude of the force in x-direction will be zero:

<u>Fₓ = 0 N</u>

For upward direction the force will be negative:

<u></u>F_y<u> = - 9.08 N</u>

<u></u>

(c)

Since the motion is in the horizontal direction. Therefore the magnitude of the force in y-direction will be zero:

<u></u>F_y<u> = 0 N</u>

<u>Fₓ = 9.08 N</u>

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