Question

A converging lens of focal length 20.0 cm forms images of an object situated at various distances. (a) If the object is placed 40.0 cm from the lens, locate the image, state whether it's real or virtual, and find its magnification. (b) Repeat the problem when the object is at 20.0 cm and (c) again when the object is 10.0 cm from the lens. (d) Follow through with ray tracing.

And Repeat the above problem, but now for a diverging lens of focal length 10.0 cm. Similarly, draw a ray trace to show the effect of a diverging lens.

Answer 1

u = object distance and v =image distance

Part 1: converging lens,

f = +20 cm

a) u = -2f = - 40 cm

image is formed at 2f on the other side.

so, v = + 40 cm

or, by formula also,

1/v = 1/u + 1/f = -1/40 + 1/20

or, v = +40 cm

m = v/u = 40 / (- 40) = -1

Image is real and inverted.

b) u = - f = -20 cm

we can see that the image is formed at infinity.

or, by formula also,

1/v = - 1/20 + 1/20

or, v = infinite

So, m = infinite

Image is real and inverted.

c) u = - 10 cm = - f/2

we can see that the image is formed at focus on the left side.

v = - f = - 20 cm

or, by formula also,

1/v = - 1/10 + 1/20

or, v = - 20 cm

m = - 20/ - 10 = + 2

Image is virtual and erect.

Part 2: Diverging Lens

f = -10 cm

a) u = - 40 cm

since, 1/v - 1/u = 1/f

or, 1/v = - 1/10 + (-1/40)

or, v = - 8 cm

m= v/u = - 8 / ( - 40) = + 0.2

Image is virtual and erect.

b) u = - 20 cm

1/v = 1/u + 1/f

or, 1/v = -1/20 - 1/10

or, v = - 20/3 = - 6.67 cm

m = v/u = - 6.67 / - 20 = + 0.33

Image is virtual and erect.

c) u = - 10 cm = - f

We can see that the image is formed at f/2 distance to the left of lens.

so, v = - 5 cm

or, by formula also,

1/v = - 1/10 - 1/10

or, v = - 5 cm

m = -5/ -10 = +0.5

Image is virtual and erect

Note: All the answers have been verified by both the ray diagram as well as from calculations.