Question

A converging lens creates the image shown in FIGURE Q34.8. Is the object distance less than the focal length f, between f and 2 f, or greater than 2 f? Explain

Short answer

The object is between f and 2f away.

Step-by-step solution

Find f:

When we have a converging lens and the image is inverted, real, and magnified (as shown in figure Q34.8), the object is backslash boxed backslash backslash text between f and 2 f away from the lens. Remember the thin lens equation?

$\frac{1}{f}=\frac{1}{s}+\frac{1}{s^{\mathrm{\prime }}}$

we get

$\frac{1}{s^{\mathrm{\prime }}}=\frac{1}{f}-\frac{1}{s}.$

Find 2f or greater thean 2f

Since$s>f$then $\frac{1}{f}>\frac{1}{s}$i.e. we see that$s^{\text{'}}$is positive meaning the image is real and on the other side of the lens (as is the case in this problem). Also, $s<2f$means that $\frac{1}{s}>\frac{1}{2f}$i.e. $\frac{1}{f}-\frac{1}{s}<\frac{1}{f}-\frac{1}{2f}=\frac{1}{2f}$. Therefore

$\frac{1}{s^{\mathrm{\prime }}}<\frac{1}{2f}$

yielding

So we got that only when f<s<2 f then$s^{\text{'}}>2f$i.e. $s^{\text{'}}>s$and the image is also magnified magnified.