The following four lenses are placed together in close contact. Find the focal length of the combination.

Lens 1- F'= +200mm
Lens 2- F'= -125mm
Lens 3- F'= -400mm
Lens 4- F'= +142.9mm

If you guys are looking for the 4 question one the answers are

A
C
B
C

For two lenses:

1/f = 1/f1 + 1/f2

where f is the focal length of the lens combination, f1 is the focal length of the first lens, and f2 is the focal length of the 2nd lens

Extending this to 4 lenses:

1/f = 1/f1 + 1/f2 + 1/f3 + 1/f4

or

1/f = 1/200 - 1/125 -1/400 + 1/142.9

1/f = 0.0014979

f = 667.6 mm

Lens 1- F'= +200mm=+0.2 m

Lens 2- F'= -125mm = - 0.125 m
Lens 3- F'= -400mm =- 0.4 m
Lens 4- F'= +142.9mm= +0.1429 m

1/F = 1/0.2 – 1/0.125 – 1/0.4 +1/0.1429 =
=5-8-2.5+7 = -3.5
F= - 0.286 m

Jenifer is right

1/F = 1/0.2 – 1/0.125 – 1/0.4 +1/0.1429 =
=5-8-2.5+7 = +1.5
F= + 0.667 m

thank you!

To find the focal length of the combination of lenses, you can use the lensmaker's formula:

1/F = (n - 1) * (1/R1 - 1/R2)

Where:
- F is the focal length of the lens
- n is the refractive index of the material the lens is made of
- R1 is the radius of curvature of the first surface of the lens
- R2 is the radius of curvature of the second surface of the lens

Since the lenses are placed in close contact, the effective focal length of the combination can be found by summing the individual focal lengths:

1/F_effective = 1/F1 + 1/F2 + 1/F3 + 1/F4

Let's calculate the effective focal length step by step:

For Lens 1: F1 = +200mm
For Lens 2: F2 = -125mm
For Lens 3: F3 = -400mm
For Lens 4: F4 = +142.9mm

Now let's substitute these values in the formula:

1/F_effective = 1/200 + 1/(-125) + 1/(-400) + 1/142.9

Calculating this expression will give us the reciprocal of the effective focal length. To find the effective focal length itself, take the reciprocal of the result:

F_effective = 1 / (1/200 + 1/(-125) + 1/(-400) + 1/142.9)

By evaluating the expression, you will get the value of the effective focal length of the combination of lenses.