Question

20 cm A diverging lens with a focal length of -10 cm is on the same axis as a converging lens with a focal length of +15 cm as illustrated. The distance between the lenses is 20 cm. An object is placed 30 cm to the left of the diverging lens. a) (10 Points) Where is the final image formed? (Submit a PDF file with a maximum size of 8 MB) Choose File No file chosen b) (10 Points) Calculate the overall magnification. (Submit PDF file with a maximum size of 8 MB) Choose File No file chosen c) (5 Points) If the diverging lens was removed so that only the converging lens is left, use ray tracing technique to determine if the image formed by the converging lens is real or virtual, and whether it is inverted or upright. (Submit a PDF file with a maximum size of 8 MB) Choose File No file chosen

Answer 1

If p is the object distance and q is the image distance, then the focal length f is given by,

$\frac{1}{f}=\frac{1}{q}-\frac{1}{p}$

The magnification of the lens is,

M D

(a) For the diverging lens, f = -10cm and p = -30cm. Therefore,

(-10) 9 (-30)

$\frac{1}{q}=\frac{1}{(-10)}+\frac{1}{(-30)}$

$\frac{1}{q}=\frac{(-30)+(-10)}{(-30)(-10)}=\frac{-40}{300}$

$q=\frac{300}{-40}=-7.5cm$

The image formed by the diverging lens is 7.5cm on the same side of the object.

For the converging lens, f = 15cm and p = -(20+7.5)=-27.5cm. Therefore,

$\frac{1}{15}=\frac{1}{q}-\frac{1}{(-27.5)}$

$\frac{1}{q}=\frac{1}{15}+\frac{1}{(-27.5)}$

$\frac{1}{q}=\frac{(-27.5)+15}{(-27.5)\times 15}=\frac{-12.5}{-412.5}$

$q=\frac{-412.5}{-12.5}=33cm$

The image formed by the converging lens is 33cm on the opposite of the object. Therefore the final image is formed 33cm to the right of the converging lens.

(b) For the diverging lens, the magnification is

$M_{1}=\frac{-7.5}{-30}=0.25$

For the converging lens, the magnification is

$M_{2}=\frac{33}{-27.5}=-1.2$

The overall magnification id,

$M=M_{1}M_{2}$

$M=0.25\times -1.2=-0.3$

The overall magnification is -0.3.

(c) If the diverging lens was removed, the new object distance for the converging lens becomes 20+30 = 50cm.

Scale X-axis = 1 unit = 10cm Y-axis 1 unit = 10cm IMAGE 27 OBJECT

The image formed by the converging lens alone is inverted and real.