Question
Suppose an object has thickness $d p$ so that it extends from object distance $p$ to $p+d p$ . Prove that the thickness $d q$ of its image is given by $\left(-q^{2} / p^{2}\right) d p,$ so that the longitudinal magnification $d q / d p=-M^{2},$ where $M$ is the lateral magnification.
Step 1
Step 1: Recall the lens formula, which relates the object distance \( p \), the image distance \( q \), and the focal length \( f \) of a lens: \[ \frac{1}{f} = \frac{1}{p} + \frac{1}{q} \] Show more…
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Suppose an object has thickness $d p$ so that it extends from object distance $p$ to $p+d p .$ Prove that the thickness $d q$ of its image is given by $\left(-q^{2} / p^{2}\right) d p .$ Then the longitudinal magnification is $d q / d p=-M^{2},$ where $M$ is the lateral magnification.
Suppose an object has thickness $d p$ so that it extends from object distance $p$ to $p+d p .$ (a) Prove that the thickness $d q$ of its image is given by $\left(-q^{2} / p^{2}\right) d p .$ (b) The longitudinal magnification of the object is $M_{\text {long }}=d q / d p .$ How is the longitudinal magnification related to the lateral magnification $M ?$
Suppose an object has thickness $d p$ so that it extends from object distance $p$ to $p+d p .$ (a) Prove that the thickness dq of its image is given by $\left(-q^{2} / p^{2}\right) d p .$ (b) The longitudinal magnification of the object is $M_{\text {lowg }}=d q / d p .$ How is the longitudinal magnification related to the lateral magnification $M ?$
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