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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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Well, first of all, one who is sufficiently educated to deal with solving
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                                                                = - 0.1755 M sin(2π t/15) .

There's the important number ... the  0.1755 M.
That's the peak acceleration.
From here, the problem is a piece-o-cake.

The net vertical force on the intrepid sailor ... the guy standing on the
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At the instant of peak acceleration, the weighing machine thinks that
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Now I'm going to wave my hands in the air a bit:

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Apparent 'gravity' = 1.08333 x real acceleration of gravity.

The increase ... the 0.08333 ... is the 'extra' acceleration that's due to
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The 'M' is what we need to find.

Divide each side by  0.1755 :          M = (0.08333 / 0.1755) G

'G' = 9.0 m/s²
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That result fills me with an overwhelming sense of no-confidence.
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