Optics/Lens problem. Is the image real or virtual?
A converging lens of focal length f1 = +22 .5cm is placed at a distance d = 60 .0cmtotheleft 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. (The thin-lens approximation may be assumed to hold.)
Is the final image real or virtual? And is it inverted or ereet?
To determine if the final image is real or virtual, and if it is inverted or erect, we can use the lens formula and the magnification formula.
1. Calculate the position of the image formed by the converging lens:
Using the lens formula:
1/f1 = 1/v1 - 1/u1
where f1 is the focal length of the converging lens, v1 is the distance of the image from the lens, and u1 is the distance of the object from the lens.
Given: f1 = +22.5 cm (converging lens)
u1 = -45 cm (object is to the left of the lens)
v1 = ?
Rearranging the formula, we have:
1/v1 = 1/f1 - 1/u1
1/v1 = 1/22.5 - 1/-45
Calculating the above expression, we get:
1/v1 = 1/22.5 + 1/45
1/v1 = (2 + 1)/45
1/v1 = 3/45
v1 = 45/3
v1 = 15 cm
The image formed by the converging lens is located 15 cm to the right of it.
2. Calculate the position of the object for the diverging lens:
Since the diverging lens is placed at a distance of 60 cm to the left of the converging lens, the distance between the object and the diverging lens is:
u2 = u1 + d
u2 = -45 + (-60)
u2 = -105 cm
3. Calculate the position of the image formed by the diverging lens:
Using the lens formula:
1/f2 = 1/v2 - 1/u2
where f2 is the focal length of the diverging lens, v2 is the distance of the image from the lens, and u2 is the distance of the object from the lens.
Given: f2 = -30.0 cm (diverging lens)
u2 = -105 cm
v2 = ?
Rearranging the formula, we have:
1/v2 = 1/f2 - 1/u2
1/v2 = 1/-30 - 1/-105
Calculating the above expression, we get:
1/v2 = -7/210
v2 = -210/7
v2 ≈ -30 cm
The image formed by the diverging lens is located approximately 30 cm to the left of it.
4. Determine if the final image is real or virtual:
Since the image formed by the diverging lens is virtual (located to the left of the lens), and the image formed by the converging lens is real (located to the right of the lens), the final image is a virtual image.
5. Determine if the final image is inverted or erect:
Since the image formed by the converging lens is inverted and the image formed by the diverging lens is also inverted, the final image is inverted.