Problem G. 800. (December 2022)

G. 800. An object is located at a certain distance from a converging lens, which forms the real, inverted image of the object of magnification \(\displaystyle N_1\). When the object is placed further from the lens, with a distance of \(\displaystyle d\) along the principal axis of the lens, the magnification becomes \(\displaystyle N_2\).

What is the focal length of the lens?

(4 pont)

Deadline expired on January 16, 2023.

Sorry, the solution is available only in Hungarian. Google translation

Megoldás. A kép valódi, ezért az eredeti \(\displaystyle t\) tárgytávolság nagyobb, mint az \(\displaystyle f\) fókusztávolság: Az

\(\displaystyle \frac1t+\frac1k=\frac1f\)

leképezési törvény szerint

\(\displaystyle k=\frac{tf}{t-f},\)

és így

\(\displaystyle N_1=\frac{k}{t}=\frac{f}{t-f},\qquad\text{azaz}\qquad \frac1{N_1}=\frac{t}{f}-1.\)

A második esetben a tárgytávolság \(\displaystyle t+d\), tehát

\(\displaystyle N_2=\frac{f}{t+d-f},\qquad\text{azaz}\qquad \frac1{N_2}=\frac{t+d}{f}-1=\frac1{N_1}+\frac{d}{f}.\)

Ezek szerint

\(\displaystyle \frac1{N_2}-\frac1{N_1}=\frac{d}{f},\)

vagyis

\(\displaystyle f=\frac{N_1N_2}{N_1-N_2}d.\)

Statistics:

34 students sent a solution.
4 points:	Antal Áron, Bencze Mátyás, Bohner Emese, Jávor Botond, Nagy 639 Csenge, Sütő Áron, Tajta Sára, Tóth Hanga Katalin, Žigo Boglárka.
3 points:	Hornok Máté.
2 points:	3 students.
1 point:	5 students.
0 point:	7 students.
Unfair, not evaluated:	8 solutionss.

Problems in Physics of KöMaL, December 2022