a) Suppose a lens can focus the image of an object exactly 5 metres away at a distance of precisely 2 centimetres. What is the focal length of the lens? (Note: focal lengths are normally measured in millimetres. Give your answer to one decimal place.)
b) Rearrange the formula to make y the subject.
c) Suppose a lens has a focal length of 50.0mm and an object is exactly 2
metres away. At what the will the object’s image be in focus?
+33615 a) Turn all the distances into millimetres and use your formula: 1/F = 1/5000 + 1/ 200
1/F = (200 + 5000)/(5000*200) or 1/F = 5200/106
Invert both sides
F = 106/5200 mm
$${\mathtt{F}} = {\frac{{{\mathtt{10}}}^{{\mathtt{6}}}}{{\mathtt{5\,200}}}} \Rightarrow {\mathtt{F}} = {\mathtt{192.307\: \!692\: \!307\: \!692\: \!307\: \!7}}$$
so F = 192.3 mm to the nearest millimetre
b) If 1/F = 1/x + 1/y then subtract 1/x from both sides to get 1/y = 1/F - 1/x
1/y = (x - F)/(x*F)
Invert both sides
y = x*F/(x - F)
3) See if you can now do this part yourself.
.
+33615 a) Turn all the distances into millimetres and use your formula: 1/F = 1/5000 + 1/ 200
1/F = (200 + 5000)/(5000*200) or 1/F = 5200/106
Invert both sides
F = 106/5200 mm
$${\mathtt{F}} = {\frac{{{\mathtt{10}}}^{{\mathtt{6}}}}{{\mathtt{5\,200}}}} \Rightarrow {\mathtt{F}} = {\mathtt{192.307\: \!692\: \!307\: \!692\: \!307\: \!7}}$$
so F = 192.3 mm to the nearest millimetre
b) If 1/F = 1/x + 1/y then subtract 1/x from both sides to get 1/y = 1/F - 1/x
1/y = (x - F)/(x*F)
Invert both sides
y = x*F/(x - F)
3) See if you can now do this part yourself.
.
