## Acalytica

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A candle is placed at a distance of $30 \mathrm{~cm}$ from a converging lens having a focal length of $10 \mathrm{~cm}$.

(a) At what distance from the lens the candle's image is formed?
(b) Is this image real or virtual?
(c) Is this image upright or inverted?
(d) Draw a ray diagram for this configuration and verify the above results.
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The known data is: The object distance $d_{o}=30 \mathrm{~cm}$ and the focal length $f=+10 \mathrm{~cm}$. All that is needed is to use the thin-lens equation and magnification formula.
(a) Solving thin-lens equation below for the unknown image distance $d_{i}$, we have
\begin{aligned} \frac{1}{f} &=\frac{1}{d}_{o}+\frac{1}{d}_{i} \\ \frac{1}{10} &=\frac{1}{30}+\frac{1}{d_{i}} \\ \Rightarrow \frac{1}{d_{i}} &=\frac{1}{10}-\frac{1}{30} \\ &=\frac{3-1}{30}=\frac{2}{30} \end{aligned}
Inverting above expression, get the image distance as $d_{i}=+15 \mathrm{~cm} .$

(b) Because the image distance was obtained positive, so the candle's image is real and formed on the backside of the lens.

In other words, the image is formed on the opposite side of the lens where the object is present. This is the definition of a real image formed by a converging (convex) lens.
(c) The magnification formula for thin-lens, which is negative of the ratio of image distance to object distance, get the orientation of the image formed.
$M=-\frac{d_{i}}{d_{o}}$
A negative magnification, $M<0$, indicates an inverted image and for $M>0$ an upright image is formed.

In this example problem, we have
$M=-\frac{d_{i}}{d_{o}}=-\frac{+15}{30}=-0.5$
A negative magnification is obtained, so the candle's image is inverted.
(d) All the above results are also obtained by a ray diagram shown below.

by Platinum (164,920 points)

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