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# thin lens formula

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1. 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.)

2. b)  Rearrange the formula to make y the subject.

3. 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?

Guest May 3, 2015

#1
+27219
+10

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.

.

Alan  May 3, 2015
#1
+27219
+10

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.

.

Alan  May 3, 2015