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*******PLEASE HELP***** A converging lens has a focal length f of 7.00 cm. Calcu

ID: 1543758 • Letter: #

Question

*******PLEASE HELP*****

A converging lens has a focal length f of 7.00 cm. Calculate the image position, s', for each of these object positions. Remember to keep a lot of sig figs during your calculation, then enter your answer rounded to THREE SIG FIGS.

(a) s = 3.7f. In other words, s = 3.7 times the numerical value of f.
cm

(b) s = 2f
cm

(c) s = 1.6f
cm

(d) s = 0.6f
cm
(e) In which of these cases will the magnification m be positive?
***** select one********

s > f

s = 2f    

s > 2f

s < f

m will never be positive

s = f

m will always be positive

Explanation / Answer

for converging lens, focal length is positive

hence f=7 cm

let object distance=u and image distance =v

in each case u=-k*f where k >0

then using lens equation:

(1/v)-(1/u)=1/f

==>(1/v)+(1/(k*f))=1/f

==>1/v=(k-1)/(k*f)

==>v=k*f/(k-1)=7*k/(k-1) cm

part a:

k=3.7

hence image distance=3.7*7/(3.7-1)=9.59 cm

part b:

k=2

s'=14.0 cm

part c:

k=1.6

s'=18.7 cm

part d:

k=0.6

s'=-10.5 cm


part e:

magnification=image distance/object distance

only in case of s<f, image will be in same side of the object

and magnification positive.

hence fourth choice is correct.