Question

A diverging lens has a focal length of 18.6 cm .

A. What is the image distance for an object distance of 37.2 cm? Answer with −1000 cm if no image is formed. What is the magnification?

B. What is the image distance for an object distance of 18.6 cm? Answer with −1000 cm if no image is formed. Answer in units of cm. What is the magnification?

C. What is the image distance for an object distance of 9.3 cm? Answer with −1000 cm if no image is formed. Answer in units of cm. What is the magnification?

Answer 1

focal length of diverging lense is -ve

f=-18.6cm

a)

object distance u=37.2cm

image distance =v

use,

1/u+1/v=1/f

1/37.2+1/v=1/-18.6

1/v=(-1/18.6)-(1/37.2)

=-(1/18.6+1/37.2)

====>

v=-(18.6*37.2)/(18.6+37.2)

v=-12.4 cm

magnification m=-v/u

=-(-12.4)/37.2

=0.333

hence image is erected,virtual

if

v=-1000cm

magnification m=-v/u

=-(-1000)/37.2

=26.88

b)

object distance u=18.6cm

image distance =v

use,

1/u+1/v=1/f

1/18.6+1/v=1/-18.6

1/v=(-1/18.6)-(1/18.6)

=-(1/18.6+1/18.6)

====>

v=-(18.6*18.6)/(18.6+18.6)

v=-9.3 cm

magnification m=-v/u

=-(-9.3)/18.6

=0.5

hence image is erected,virtual

if

v=-1000cm

magnification m=-v/u

=-(-1000)/18.6

=53.76

c)

object distance u=9.3cm

image distance =v

use,

1/u+1/v=1/f

1/9.3+1/v=1/-18.6

1/v=(-1/18.6)-(1/9.3)

=-(1/18.6+1/9.3)

====>

v=-(18.6*9.3)/(18.6+9.3)

v=-6.2 cm

magnification m=-v/u

=-(-6.2)/9.3

=0.666

hence image is erected,virtual

if

v=-1000cm

magnification m=-v/u

=-(-1000)/9.3

=107.52