A converging lens of focal length f1 = +22 .5cm is placed at a distance d = 60 .0cm to the left of a diverging lens of focal length f2 = −30.0cm. An object is placed on the common optical axis of the two lenses with its base 45.0cm to the left of the converging lens.

(a) Calculate the location of the final image and its overall magnification with respect to the object.

Lens A

1/u1 + 1/v1 = 1/FA

1/45.0cm + 1/v1 = 1/22.5cm

v1 = 45.0cm

This value is positive so the image is real.

Lens B

u2 = d−v1 = 60.0cm−45.0cm = 15.0cm

Again this value is positive so the image is real.

1/u2 + 1/v2 = 1/FB

1/15.0cm + 1/v2 = 1/30.0cm

v2 = −30.0cm

This value is negative so it is a virtual image.

Magnification
m = v1/u1 × v2/ u2 m = 45.0cm/45.0cm × 30.0cm/15.0cm

m = 2×

The final image is located 30.0cm to the right of the diverging lens and has an overall magnification of 2×.

m = 2

And that concludes the calculations. The overall magnification is 2, which means the image is twice the size of the object. But hey, at least it's not twice as scary as a clown!

m = 2 x 2 = 4

To calculate the location of the final image, we first use the lens formula for the converging lens (Lens A):

1/u1 + 1/v1 = 1/f1

where u1 is the object distance, v1 is the image distance, and f1 is the focal length of Lens A.

In this case, the object distance u1 is 45.0 cm (as given in the question) and the focal length f1 is +22.5 cm (positive because it's a converging lens).

Plugging in these values, we get:

1/45.0 cm + 1/v1 = 1/22.5 cm

Simplifying the equation, we find:

1/v1 = 1/22.5 cm - 1/45.0 cm

v1 = 45.0 cm

Since the calculated value of v1 is positive, the image is real.

Next, we move on to the diverging lens (Lens B). The object distance for Lens B, u2, is the difference between the distance d (distance between the lenses) and the previously calculated image distance v1:

u2 = d - v1 = 60.0 cm - 45.0 cm = 15.0 cm

Again, since the calculated value of u2 is positive, the image formed by Lens B is real.

Now we use the lens formula for the diverging lens (Lens B):

1/u2 + 1/v2 = 1/f2

where u2 is the object distance (15.0 cm), v2 is the image distance, and f2 is the focal length of Lens B.

In this case, the focal length f2 is -30.0 cm (negative because it's a diverging lens).

Plugging in the values, we have:

1/15.0 cm + 1/v2 = 1/30.0 cm

Simplifying the equation, we find:

1/v2 = 1/30.0 cm - 1/15.0 cm

v2 = -30.0 cm

Since the calculated value of v2 is negative, the image formed by Lens B is virtual.

Finally, to calculate the overall magnification, we use the formula:

m = (v1/u1) × (v2/u2)

In this case, v1 is 45.0 cm, u1 is 45.0 cm, v2 is -30.0 cm, and u2 is 15.0 cm.

Plugging in these values, we obtain:

m = (45.0 cm/45.0 cm) × (30.0 cm/15.0 cm)

m = 2

Therefore, the overall magnification of the final image with respect to the object is 2.