A candle flame 1.6 cm high is imaged in a ball bearing of diameter 0.4 cm . If the ball bearing is 20 cm away from the flame, determine the location and the height of the image.

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Quantrex Academy · Accepted Answer

Correct answer: C. 1.0 mm inside the ball bearing, 0.08 mm

The ball bearing acts as a convex mirror.

Radius of curvature

R = \dfrac{0.4}{2} = 0.2 \text{ cm}

.

Focal length

f = \dfrac{R}{2} = +0.1 \text{ cm}

.

Object distance

u = -20 \text{ cm}

.

Using the mirror formula:

\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}

\dfrac{1}{v} - \dfrac{1}{20} = \dfrac{1}{0.1} = 10

\dfrac{1}{v} = 10 + \dfrac{1}{20} = \dfrac{201}{20}

v = \dfrac{20}{201} \text{ cm} \approx 0.1 \text{ cm} = 1.0 \text{ mm}

Since

v > 0

, the image is virtual and formed

1.0 \text{ mm}

inside the ball bearing.

Magnification

m = -\dfrac{v}{u}

m = -\dfrac{20/201}{-20} = \dfrac{1}{201}

Height of the image

h_i = m \times h_o

h_i = \dfrac{1}{201} \times 1.6 \text{ cm} \approx \dfrac{1.6}{200} \text{ cm} = 0.008 \text{ cm} = 0.08 \text{ mm}

A candle flame 1.6 cm high is imaged in a ball bearing of diameter...

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