New solution

Luisito edited Jun 12, 2026, 4:45 PM

View solution History

revision #18694 newer →

en/13.1.17.md +27 −0

		@@ -0,0 +1,27 @@
		+	### Statement
		+
		+	$13.1.17.$ [Insert the problem statement]
		+
		+	### Solution
		+
		+	Firstly, the magnification of the object is given by:\
		+	$k = \frac{s'}{s}$ (1)\
		+	where $s'$ is the image-mirror distance and $s$ is the object-mirror distance, with $k$ = 3.\
		+	And the relation for these distances is\
		+	$\frac{1}{s} +\frac{1}{s'} = \frac{1}{f}$ (2)\
		+	When object is moved a distance $d$ away from the mirror, the magnification changes\
		+	$m = \frac{s''}{s+d}$ (3)\
		+	with $m$ = 0.5.\
		+	And the relation for these distances is\
		+	$\frac{1}{s+d}+\frac{1}{s''}=\frac{1}{f}$ (4)\
		+	From (1) and (2), eliminating term $s'$,\
		+	$\frac{s}{f} = 1+\frac{1}{k}$ (5)\
		+	From (3) and (4), eliminating term $s''$,
		+	$\frac{s}{f}=1+\frac{1}{m}-\frac{d}{f}$ (6)\
		+	Equaling (5) and (6), and separating $f$,\
		+	$f = \frac{mkd}{k-m} = 48\;\rm{cm}$
		+
		+
		+	#### Answer
		+
		+	[Insert a concise answer or boxed result]

	@@ -0,0 +1,27 @@
			### Statement

			$13.1.17.$ [Insert the problem statement]

			### Solution

			Firstly, the magnification of the object is given by:\
			$k = \frac{s'}{s}$ (1)\
			where $s'$ is the image-mirror distance and $s$ is the object-mirror distance, with $k$ = 3.\
			And the relation for these distances is\
			$\frac{1}{s} +\frac{1}{s'} = \frac{1}{f}$ (2)\
			When object is moved a distance $d$ away from the mirror, the magnification changes\
			$m = \frac{s''}{s+d}$ (3)\
			with $m$ = 0.5.\
			And the relation for these distances is\
			$\frac{1}{s+d}+\frac{1}{s''}=\frac{1}{f}$ (4)\
			From (1) and (2), eliminating term $s'$,\
			$\frac{s}{f} = 1+\frac{1}{k}$ (5)\
			From (3) and (4), eliminating term $s''$,
			$\frac{s}{f}=1+\frac{1}{m}-\frac{d}{f}$ (6)\
			Equaling (5) and (6), and separating $f$,\
			$f = \frac{mkd}{k-m} = 48\;\rm{cm}$


			#### Answer

			[Insert a concise answer or boxed result]