Using the Lens or Mirror Equation, , where s is the distance from the object to

Question

Using the Lens or Mirror Equation, , where s is the distance from the object to the instrument, s’ is the distance from the image to the instrument and f is the distance from the focal point to the instrument. Show work on separate paper. In each case tell whether the image is real, virtual, larger than or smaller than the object and inverted or erect. Do the same thing with the objects. Also, find the image’s magnification from O/I=s/s'. Use the absolute value of s/s'. Find : 1) (1/12)+(1/6)=(1/f), find f. 2) s=-5, f=20. find s'. 3) (1/22)+(1/s')=(1/12), find s`. 4) (1/s)+(1/6)=(1/12), find s. 5) s`=16, f=12, find s. 6) s=-12.6, s`=6.4, find f. 7) (1/-12)+(1/6)=(1/f), find f. 8) (1/5)+(1/s`)=(1/-20), find s`. 9) (1/22)+(1/s`)=(1/-12), find s`. 10) s=2.4, s`=3.6, find f.

Explanation / Answer

Find : 1) (1/12)+(1/6)=(1/f), find f.
1/f = 1/12 + 1/6
1/f = (1+2)/12
f = 12/3 = 4
Answer: 4

2) s=-5, f=20. find s'.
1/f = 1/s + 1/s'
1/20 = -1/5 + 1/s'
1/s' = 1/20 + 1/5
1/s' = (1+4)/20
s' = 20/5
     = 4

3) (1/22)+(1/s')=(1/12), find s`.
(1/22)+(1/s')=(1/12),
1/s' = 1/12 - 1/22
1/s' = (11 - 6)/132
s' = 132/5
     = 26.4

4) (1/s)+(1/6)=(1/12), find s.
(1/s)+(1/6)=(1/12)
1/s = 1/12 - 1/6
1/s = (1-2)/12
s = -12
Answer: -12

I am allowed to answer only 4 parts at a time

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