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VladimirAG [237]
11 months ago
13

what is the thickness of the central portion of a thin conveying lens can be determined very accurately by using (a) vernier cal

liper (b) micro meter screw guage (c) telescope ( d) microscope​
Physics
1 answer:
beks73 [17]11 months ago
8 0

Option B The thickness of the central portion of a thin conveying lens can be determined very accurately by using a micrometer screw gauge.

<h3>What can be measured using a micrometer screw gauge?</h3>

One micrometer of thickness can be measured with a micron micrometre screw gauge. A Use of Micrometer Screw Gauge as like example Upon turning the screw of the micrometer screw gauge four times, a 2 mm space is covered.

<h3>What purposes does a micrometer serve?</h3>

A tool known as a micrometer is used to measure solid objects’ lengths, thicknesses, and other dimensions precisely and linearly.

<h3>What is the micrometer screw gauge’s SI unit?</h3>

The SI symbol m is also known as a micron, which is an SI-derived unit of length equaling 1106 meters, where 106 is the SI standard prefix for the prefix “micro-.” A micrometer is one-millionth of a meter.

To know more about screw gauges, visit:

brainly.com/question/4704005

#SPJ13

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1. A small light bulb is shining light on a basketball (diameter is 23 cm or 9 inches). The light bulb is 3 m from the closest s
siniylev [52]

Answer:

The size (diameter) of the basketball's shadow on the wall is approximately 53.38 cm

Explanation:

The given parameters of the basketball are;

The diameter of the basketball = 23 cm (9 inches)

The distance of the light bulb from the closest side of the basketball = 3 m

The distance from the ball to the wall = 4 m

The distance from the light source to the center of the ball, d = 3 m + 0.23/2 m = 3.115 m

The angle the light ray makes with the edge of the ball, θ = arctan(0.115/3.115)

Therefore, the ratio of the shadow width divided by 2 to the distance from the light from the wall = 0.115/3.115

The distance from the light from the wall = 3 m + 4 m + 0.23 m = 7.23 m

Therefore;

((The width of the shadow)/2)/(The distance from the light from the wall) = 0.115/3.115

∴ ((The width of the shadow)/2)/(7.23 m) = 0.115/3.115

((The width of the shadow)/2) = 7.23 m × 0.115/3.115 = 16629/62300 m ≈ 0.2669 m = 26.69 cm

The width (diameter) of the shadow on the wall = 2 × 16629/62300 m ≈ 0.5338 m = 53.38 cm

The size (diameter) of the basketball's shadow on the wall ≈ 53.38 cm

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3 years ago
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