Optics(Converging lens and Diverging lens)

[BSElectrician]

Homework Statement

http://www.imageurlhost.com/images/r0its2lyef7v3s0ppms2.png

Homework Equations

Mirror Equation
(1/q) + (1/p) = (1/f) = (2/r)
Magnification
M = (q/p)

for two lenses
M1 & M2
Total magnification = M1*M2

The Attempt at a Solution

we have p1 = 40cm and f1 = 30cm
then the image distance of lens 1 will be q1 = 120

then we have diverging lens 2
with focal length f2 = 20cm

then I am stuck i don't know what to do with the object distance of the diverging lens
i also don't know what to do with the fact that the object is 2cm high

 

[grzz]

The image of the converging lens acts as a VIRTUAL object for the diverging lens.

 

[BSElectrician]

The image of the converging lens acts as a VIRTUAL object for the diverging lens.

is it between the diverging and converging lens?

this is how i do to find the image distance of the divergng lens
between C lens and D lens (110cm)
but there is an image distance q1 from the C lens (120cm)
the object distance from the D lens p2 might be (110cm - 120cm)(1/q2) + (1/(110-120)) = (1/20)
q2 = 6.67
then i can find the magnification of the Divirging lens
M2 = q2/p2
total
M = M1M2

whats the use of that 2cm thing above the object?

 

[grzz]

If the formula (1/u) + (1/v) = 1/f is to be used then the focal length of the diverging lens is taken as -20.

 

[asteriatic]

I would approach this problem by first understanding the principles of optics and the behavior of converging and diverging lenses. A converging lens, also known as a convex lens, has a positive focal length and is thicker in the middle than at the edges. It causes parallel rays of light to converge at a focal point, forming a real image. On the other hand, a diverging lens, also known as a concave lens, has a negative focal length and is thinner in the middle than at the edges. It causes parallel rays of light to diverge, making them appear to come from a virtual focal point.

In this problem, we are given the object distance (p) and focal length (f) of the first lens, and we need to find the image distance (q) of the first lens. We can use the mirror equation (1/q + 1/p = 1/f) to find this value. Since the object distance (p) is positive (40cm), the image distance (q) should also be positive. Therefore, the image formed by the first lens will be a real image, located at a distance of 120cm from the lens.

Now, we need to consider the second lens. Since it is a diverging lens, the image formed by the first lens will act as the object for the second lens. This means that the object distance (p) for the second lens will be 120cm. Since we are given the focal length (f) of the second lens, we can use the mirror equation again to find the image distance (q) of the second lens. However, we also need to take into account the fact that the object (the image formed by the first lens) is 2cm high. This means that the image formed by the second lens will also be 2cm high, and we can use the magnification equation (M = q/p) to find the image distance (q) of the second lens.

Once we have the image distance (q) of the second lens, we can use the total magnification equation (M1*M2) to find the total magnification of the two lenses. This will give us the final image size and location.

In summary, as a scientist, I would approach this problem by understanding the principles of optics and using the appropriate equations to solve for the image distance of the first lens, taking into account the object distance and height for