A converging lens (f1 = 24.0 cm) is located 56.0 cm to the left of a diverging lens (f2 = -28.0 cm). An object is placed to the left of the converging lens, and the final image produced by the two-lens combination lies 19.7 cm to the left of the diverging lens.
How far is the object from the converging lens?
I have to be honest, when using the lens equations for multilens images, I nearly always mess up the signs. So what I do, is do a ray diagram of the first lens, then over sketch it with the image of the first lens becoming the object of the second lens, then eamining the final object. Of course, it is not accurate, and can't be used (my sketches are "engineering" sketches), but it then allow me to examine the results of the two lens equations to make certain it makes sense, and is right.
To find the distance of the object from the converging lens, we can use the lens equation:
1/f = 1/di - 1/do
where:
f = focal length of the lens
di = image distance
do = object distance
Let's assume the object distance from the converging lens is "x".
From the given information, we know:
f1 = 24.0 cm (focal length of the converging lens)
di = -19.7 cm (image distance from the diverging lens)
f2 = -28.0 cm (focal length of the diverging lens)
First, let's find the image distance produced by the converging lens:
1/f1 = 1/di - 1/do
Substituting the given values:
1/24.0 = 1/-19.7 - 1/x
Simplifying the expression:
1/24.0 = (-1/19.7) - 1/x
Now, let's find the common denominator:
1/24.0 = (-1/19.7) * (19.7/19.7) - 1/x
1/24.0 = -19.7/19.7 - 1/x
1/24.0 = (-19.7 - 1/x) / 19.7
To isolate x, we can cross multiply:
x = (19.7 * 24.0) / (-19.7 - 1/x)
Simplifying further:
x = 474.8 / (-19.7 - 1/x)
Now, let's solve this equation to find the value of x:
x (-19.7 - 1/x) = 474.8
-19.7x - 1 = 474.8
-19.7x = 475.8
x = 475.8 / -19.7
x ≈ -24.2 cm
Therefore, the object is approximately 24.2 cm to the left of the converging lens.