Question

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, find the location and the height of the image:
A. $1.0\;mm$ inside the ball bearing, $0.08\;mm$
B. $1.0\;mm$ outside the ball bearing, $0.06\;cm$
C.$2\;m$ inside the ball bearing, $0.08\;cm$
D. $1\;m$ outside the ball bearing, $0.05\;mm$

Answer 1

Hint: The mirror formula is the relationship between the distance of an object $u$ , distance of image $v$ and the focal length of the lens $f$. This law can be used for both concave and convex mirrors with appropriate sign conventions. This sum is similar to an object placed in front of a spherical mirror. Hence using spherical mirror formula, we can find the image distance. And using the magnification formula, we can find the height of the image.

Formula used:
$\dfrac{1}{u}+\dfrac{1}{v}=\dfrac{1}{f}$ and $m=\dfrac{H_{i}}{H_{o}}=-\dfrac{v}{u}$

Complete step by step answer:
Let us consider the ball bearing as a spherical mirror whose radius of curvature is $R$. Here, it is given that the diameter of the ball bearing is $0.4\;cm$, then we can say that the radius of curvature $R=0.2cm$
Also if the distance between the flame and ball bearing is $u$, then we have $u=20cm$ and the height of the candle flame is $H_{0}=1.6cm$
From mirror formula we know that $\dfrac{1}{u}+\dfrac{1}{v}=\dfrac{1}{f}$
But we also know that $R=2f$
$\implies f=\dfrac{R}{2}$
Then replacing, we get $\dfrac{1}{u}+\dfrac{1}{v}=\dfrac{2}{R}$
Substituting for $u$ and $R$, we get, $\dfrac{1}{v}=\dfrac{2}{0.2}-\dfrac{1}{-20}=\dfrac{1}{0.1}$
$\implies v=0.1cm$
We also know that magnification, $m=\dfrac{H_{i}}{H_{o}}=-\dfrac{v}{u}$
Substituting we get $\dfrac{H_{i}}{1.6}=-\dfrac{0.1}{-20}$
$\implies H_{i}=\dfrac{0.16}{20}=0.008cm$
Thus the height of the image is $0.008\;cm$ and the image is a distance $0.1\;cm$
Since the image distance $v=0.08mm$ is less than the radius of curvature $R=0.2cm$ of the ball bearing, we can say that the image is inside the ball bearing.
Hence the answer is A. $1.0\;mm$ inside the ball bearing, $0.08\;mm$

Note:
Magnification equation is used which states \[\text{M=}\dfrac{\text{Height}\ \text{of}\ \text{image}}{\text{Height}\ \text{of}\ \text{object}}\text{=-}\dfrac{\text{distance}\ \text{of}\ \text{image}}{\text{distance}\ \text{of}\ \text{object}}\]if $M=+$ then the image is magnified and if $M=-$ then the image is diminished. Here we have two cases, where the image produced is either a real or virtual image depending on where the object is placed.