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Diano4ka-milaya [45]
3 years ago
8

An amateur astronomer grinds a double convex lens whose surfaces have radii of curvature of 40 cm and 60 cm. The glass has an in

dex of refraction of 1.54. What is the focal length of this lens in air?
Physics
1 answer:
DedPeter [7]3 years ago
6 0

Answer:

44.4cm

Explanation:

glass has an index of refraction .n = 1.54

radii of curvature of 40 cm R1 = 40 by

radii of curvature of 600 cm R2 = 60

Now, by lens maker formula

1/f = (n - 1) (1/R1 - 1/R2)

Putting in the given values for n = 1.54 , we get f = 22.2

\frac{1}{f} = (1.54 -1) (\frac{1}{40cm} -\frac{1}{(-60cm)} )

\frac{1}{f} = 0.0225

f = 1 / 0.0225

f = 44.4cm

so, focal length in air will be  = 44.4 cm

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we can set the following proportion:

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The continuity equation is given as;

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A 20-foot ladder is leaning against the wall. If the base of the ladder is sliding away from the wall at the rate of 3 feet per
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Answer:

<u>6.87 ft/s</u> is the rate at which the top of ladder slides down.

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Now, from triangle ABC,

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