A light source (object) and a screen are placed 100 cm aparton an optical bench.

Question

A light source (object) and a screen are placed 100 cm aparton an optical bench. A sharp image of the light source is obtainedon the screen when a lens is positioned 20 cm from the screen. Whatis the focal length of the lens? I don't know which formula to use because I tried 1/f = 1/p +1/q and that answer was wrong and there is no index of refractionso I can't use the lensmakers equation, so I don't really know whatformula to use unless I'm misunderstanding something. Please help me? A light source (object) and a screen are placed 100 cm aparton an optical bench. A sharp image of the light source is obtainedon the screen when a lens is positioned 20 cm from the screen. Whatis the focal length of the lens? I don't know which formula to use because I tried 1/f = 1/p +1/q and that answer was wrong and there is no index of refractionso I can't use the lensmakers equation, so I don't really know whatformula to use unless I'm misunderstanding something. Please help me?

Explanation / Answer

A light source (object) and a screen are placed 100 cm aparton an optical bench. A sharp image of the light source is obtainedon the screen when a lens is positioned 20 cm from the screen. Whatis the focal length of the lens? I don't know which formula to use because I tried 1/f = 1/p +1/q and that answer was wrong and there is no index of refractionso I can't use the lensmakers equation, so I don't really know whatformula to use unless I'm misunderstanding something. Please help me?
from the data u=80cm v=20cm using f=uv/u+v=16cm

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.