New solution

Valter edited
View solution History
revision #20040 newer →
en/13.3.6.md +94 −0
@@ -0,0 +1,94 @@
+### Statement
+
+$13.3.6.$ [Insert the problem statement]
+
+### Solution
+
+### Statement
+
+$13.3.6.$ An object in the form of a segment of length $l$ is located along the optical axis of a converging lens with focal length $f$, which produces a real image of all its points. The middle of the segment is located at a distance $a$ from the lens. Determine the longitudinal magnification of the object.
+
+![For problem $13.3.6$|351x143, 50%](../../img/13.3.6/13.3.6.png)
+
+### Solution
+
+According to the figure, the distance from the left end of the object to the lens will be:
+
+$$
+a_{1}=a+\frac{l}{2} \tag{1}
+$$
+
+The distance from the right end of the object to the lens will be:
+
+$$
+a_{2}=a-\frac{l}{2} \tag{2}
+$$
+
+Using the thin lens formula, we find the distances between the lens and the images of these ends of the object:
+
+$$
+\frac{1}{a_{1}}+\frac{1}{b_{1}}=\frac{1}{f}
+$$
+$$
+b_{1}=\frac{a_{1}f}{a_{1}-f} \tag{3}
+$$
+
+Similarly for $b_{2}$:
+
+$$
+b_{2}=\frac{a_{2}f}{a_{2}-f} \tag{4}
+$$
+
+Since $a_{1}> a_{2}$, then $b_{2}> b_{1}$.
+
+The longitudinal magnification of the object, by definition, is:
+
+$$
+k=\frac{b_{2}-b_{1}}{l} \tag{5}
+$$
+
+Substituting (1) into (3), we find $b_{1}$:
+
+$$
+b_{1}=\frac{(a+\frac{l}{2})f}{(a-f)+\frac{l}{2}}
+$$
+
+Substituting (2) into (4), we find $b_{2}$:
+
+$$
+b_{2}=\frac{(a-\frac{l}{2})f}{(a-f)-\frac{l}{2}}
+$$
+
+Substituting the expressions for $b_{1}$ and $b_{2}$ into formula (5):
+
+$$
+k=\frac{1}{l}\cdot \left( \frac{(a-\frac{l}{2})f}{(a-f)-\frac{l}{2}}-\frac{(a+\frac{l}{2})f}{(a-f)+\frac{l}{2}} \right)
+$$
+
+Finding a common denominator and simplifying:
+
+$$
+k=\frac{f}{l}\cdot \frac{(a-\frac{l}{2})(a+\frac{l}{2})-f(a-\frac{l}{2})-(a-\frac{l}{2})(a+\frac{l}{2})+f(a+\frac{l}{2})}{(a-f)^{2}-\frac{l^{2}}{4}}
+$$
+$$
+k=\frac{f}{l}\cdot \frac{-f(a-\frac{l}{2})+f(a+\frac{l}{2})}{(a-f)^{2}-\frac{l^{2}}{4}}
+$$
+$$
+k=\frac{f}{l}\cdot \frac{fl}{(a-f)^{2}-\frac{l^{2}}{4}}
+$$
+
+Finally:
+
+$$
+k=\frac{f^{2}}{(a-f)^{2}-\frac{l^{2}}{4}}
+$$
+
+#### Answer
+
+$$
+k=\frac{f^{2}}{(a-f)^{2}-\frac{l^{2}}{4}}
+$$
+
+#### Answer
+
+[Insert a concise answer or boxed result]